How to Identify an Impossible Projectile Answer in HSC Physics

Imagine rolling a ball horizontally off a 5.0 m balcony at 8.0 m/s. A student uses the projectile flight-time formula \(T=2u_y/g\). Since the ball was launched horizontally, \(u_y=0\), so their calculator gives \(T=0\) seconds and a horizontal range of zero.

That answer is impossible. The ball still has to fall 5.0 m before reaching the ground. The mistake is using a formula that assumes the projectile lands at its launch height.

So, how do you identify an impossible projectile answer in HSC Physics? Check the horizontal and vertical motion separately, make sure both use the same time, and test whether the calculated position agrees with the actual landing height.

For the balcony example, the correct flight time is

\[ t=\sqrt{\frac{2h}{g}} =\sqrt{\frac{2(5.0)}{9.8}} =1.01\text{ s} \]

The horizontal distance is therefore

\[ x=8.0(1.01)=8.1\text{ m} \]

The ball travels for 1.01 seconds and covers approximately 8.1 m horizontally. Nothing about a zero initial vertical velocity prevents it from falling.

01The first test: separate horizontal and vertical motion

A projectile follows two motions simultaneously.

• Horizontally, it moves at constant velocity if air resistance is negligible.
• Vertically, it accelerates downwards because of gravity.

Think of the projectile as one object being tracked by two rulers: one measures its horizontal position, and the other measures its vertical position.

The important detail is that both rulers describe the same object at the same instant. They do not have separate clocks.

Set up the equations correctly

For a projectile launched at speed \(u\), at an angle \(\theta\) above the horizontal, resolve its initial velocity into two components:

\[ u_x=u\cos\theta \]

\[ u_y=u\sin\theta \]

Where:

• \(u_x\) is the initial horizontal velocity, in m/s.
• \(u_y\) is the initial vertical velocity, in m/s.
• \(u\) is the initial launch speed, in m/s.
• \(\theta\) is the launch angle above the horizontal.

Take rightwards as positive horizontally and upwards as positive vertically.

Ignoring air resistance:

\[ a_x=0 \]

\[ a_y=-g=-9.8\text{ m s}^{-2} \]

The horizontal displacement from the launch point is therefore

\[ \boxed{x=u_xt} \]

The vertical displacement from the launch point is

\[ \boxed{\Delta y=u_yt-\frac12gt^2} \]

These are the two equations you should return to whenever an answer seems suspicious.

The corresponding velocity equations are

\[ v_x=u_x \]

\[ v_y=u_y-gt \]

Notice that horizontal velocity remains constant, while vertical velocity changes continuously.

Five warning signs of an impossible answer

There is another useful check: at the highest point, the vertical velocity is zero, but the horizontal velocity is not.

Unless the projectile was launched vertically, its speed at the highest point cannot be zero.

02The biggest trap: confusing level-ground and elevated launches

This is one of the easiest ways to produce an answer that looks mathematically reasonable but describes the wrong physical situation.

There are two different cases. Keep them separate.

Case 1: The projectile lands at its launch height

Imagine kicking a football from ground level and having it land back on the same level ground.

Its initial and final vertical positions are equal, so

\[ \Delta y=0 \]

Substitute this into the vertical displacement equation:

\[ 0=u_yt-\frac12gt^2 \]

Factorise:

\[ 0=t\left(u_y-\frac12gt\right) \]

There are two solutions:

• \(t=0\), the instant of launch.
• \(t=2u_y/g\), the later time when the projectile returns to its original height.

Therefore, the total flight time is

\[ \boxed{T=\frac{2u_y}{g}} \]

The horizontal range is

\[ R=u_xT \]

Substituting the velocity components gives

\[ \boxed{R=\frac{u^2\sin(2\theta)}{g}} \]

These familiar formulas require the projectile to land at the same height from which it was launched, with negligible air resistance and approximately constant gravity.

They are not universal projectile formulas.

Case 2: The projectile lands at a different height

Now imagine throwing a ball from a balcony towards the ground below.

The final vertical position is lower than the initial position. Therefore,

\[ \Delta y\neq0 \]

You must use the full vertical displacement equation:

\[ \Delta y=u_yt-\frac12gt^2 \]

Rearranging into quadratic form gives

\[ \frac12gt^2-u_yt+\Delta y=0 \]

Solving for time:

\[ \boxed{ t=\frac{u_y\pm\sqrt{u_y^2-2g\Delta y}}{g} } \]

This equation handles different landing heights.

For an upward launch:

• If the landing point is below the launch point, there is one positive flight-time solution.
• If the landing point is above the launch point but below maximum height, there are two positive solutions.
• If the landing point is exactly at maximum height, there is one solution.
• If the landing point is above maximum height, there is no real solution. The projectile cannot reach it.

The maximum vertical displacement above the launch point is

\[ \boxed{H=\frac{u_y^2}{2g}} \]

This gives you an immediate physical limit to check.

A special case: horizontal launch from a height

For a horizontal launch,

\[ u_y=0 \]

If the projectile starts a height \(h\) above the ground, its vertical displacement at impact is

\[ \Delta y=-h \]

Therefore,

\[ -h=-\frac12gt^2 \]

Rearranging:

\[ \boxed{t=\sqrt{\frac{2h}{g}}} \]

The horizontal distance is then

\[ \boxed{R=u_x\sqrt{\frac{2h}{g}}} \]

This is a distinct launch-height case, not another version of the level-ground range formula.

One particularly useful distinction is the meaning of \(y\).

If your origin is at the launch point, the vertical equation calculates displacement relative to that point.

If your origin is at ground level and the launch height is \(h\), the absolute vertical position is

\[ y=h+u_yt-\frac12gt^2 \]

Both coordinate choices work. Just don’t use the launch-point equation and then set \(y=0\) for ground impact when the ground is actually below the launch point.

For a closer explanation of the assumptions behind this model, see Projectile Motion Assumptions in HSC Physics Explained.

03How an experiment can expose an impossible answer

Suppose two students calculate different trajectories for the same launched ball. How could you decide which answer is consistent with the actual motion?

A video-tracking experiment provides a useful test.

Setting up the experiment

Imagine a ball launcher positioned on the left, firing a ball across a clear space.

Place a calibrated vertical grid behind the projectile’s path, in the same plane as the motion. Position a fixed camera to the side, with its lens perpendicular to that plane.

The camera records the ball’s position at regular time intervals.

For each video frame, record:

• The horizontal position \(x\).
• The vertical displacement \(\Delta y\).
• The elapsed time \(t\), calculated from the frame number and recording frequency.

What should the experiment predict?

If the projectile follows the ideal model, equal time intervals should produce equal horizontal displacements.

For example, if its horizontal velocity is 8.0 m/s and the camera records 50 frames per second, the time between frames is

\[ \Delta t=\frac{1}{50}=0.020\text{ s} \]

The horizontal distance between successive frames should be

\[ \Delta x=8.0(0.020) \]

\[ \boxed{\Delta x=0.16\text{ m}} \]

The vertical motion behaves differently.

Its vertical velocity changes by

\[ \Delta v_y=a_y\Delta t \]

\[ =(-9.8)(0.020) \]

\[ \boxed{\Delta v_y=-0.196\text{ m s}^{-1}} \]

So the horizontal spacing remains constant, while the vertical spacing changes.

That difference is exactly what the two-component model predicts.

What could go wrong with the experiment?

A camera held at an angle to the plane of motion can distort the apparent distances. A measurement grid placed in a different plane can introduce parallax.

An initial design using separate timing records for horizontal and vertical positions also creates a problem. The two readings might not describe the same instant.

The revised setup uses:

• A fixed camera perpendicular to the plane of motion.
• A calibrated grid in the plane being measured.
• One video recording and one shared frame-based time measurement.

This makes it possible to compare both coordinates at exactly the same time.

The numbers in this example are illustrative, not measurements from an actual experiment. In a real investigation, measurement uncertainty must also be considered. A small difference from a theoretical prediction does not automatically prove the model is wrong.

The important conclusion is that a candidate answer must agree with both components of motion at the same instant.

04Worked examples: diagnosing suspicious answers

05Questions and solutions

The following questions move from straightforward calculations to more demanding consistency checks. In each case, assume negligible air resistance and constant gravitational acceleration unless stated otherwise.

Question 1

A ball is launched horizontally from a 20.0 m-high cliff at 6.0 m/s.

A student argues that because the initial vertical velocity is zero, the projectile has a flight time of zero.

Calculate the actual flight time and horizontal distance travelled before impact. Use \(g=9.8\text{ m s}^{-2}\).

Solution 1

The ball is in flight for 2.02 s and travels 12.1 m horizontally. Its zero initial vertical velocity does not prevent it from accelerating downwards.

Step 1: Identify the vertical displacement.

\[ \Delta y=-20.0\text{ m} \]

The initial vertical velocity is

\[ u_y=0 \]

Step 2: Use the vertical displacement equation.

\[ \Delta y=u_yt-\frac12gt^2 \]

Substitute:

\[ -20.0=0-\frac12(9.8)t^2 \]

\[ 20.0=4.9t^2 \]

\[ t^2=\frac{20.0}{4.9} \]

\[ \boxed{t=2.02\text{ s}} \]

We choose the positive root because the ball reaches the ground after launch.

Step 3: Calculate horizontal distance.

\[ x=u_xt \]

\[ x=6.0(2.02) \]

\[ \boxed{x=12.1\text{ m}} \]

The tempting mistake is to assume that zero initial vertical velocity means zero vertical displacement.

In fact, gravity produces vertical velocity immediately after launch. The ball begins falling even though it initially moves horizontally.

Question 2

A ball is launched from a balcony 15.0 m above the ground at 12.0 m/s, at \(40^\circ\) above the horizontal.

A solid vertical wall stands 20.0 m horizontally from the launch point. The wall extends from ground level to a height of 9.0 m.

A student calculates the flight time using \(T=2u_y/g\) and concludes that the ball lands before reaching the wall.

Determine whether the ball clears the wall or collides with it. Explain why the student’s reasoning is incorrect.

Solution 2

The ball collides with the wall. When it reaches the wall, its height is approximately 8.59 m, which is below the wall’s 9.0 m top.

Step 1: Resolve the initial velocity.

\[ u_x=12.0\cos40^\circ =9.19\text{ m s}^{-1} \]

\[ u_y=12.0\sin40^\circ =7.71\text{ m s}^{-1} \]

Step 2: Calculate the time taken to reach the wall.

Use horizontal motion:

\[ x=u_xt \]

Rearranging:

\[ t=\frac{x}{u_x} \]

\[ t=\frac{20.0}{9.19} \]

\[ \boxed{t=2.18\text{ s}} \]

This is the time at which the ball reaches the wall’s horizontal position.

Step 3: Calculate its vertical displacement at this time.

\[ \Delta y=u_yt-\frac12gt^2 \]

Substitute:

\[ \Delta y=7.71(2.18)-4.9(2.18)^2 \]

\[ \Delta y\approx-6.41\text{ m} \]

The ball has moved 6.41 m downwards relative to its launch point.

Step 4: Calculate its height above ground.

The launch height is 15.0 m, so

\[ y=15.0-6.41 \]

\[ \boxed{y=8.59\text{ m}} \]

Compare this with the wall height:

\[ 8.59\text{ m}<9.0\text{ m} \]

The ball reaches the wall below its top and therefore collides with it.

The student’s error was treating the time to return to the launch height as the time to reach the ground.

Their calculation gives

\[ T=\frac{2(7.71)}{9.8} =1.57\text{ s} \]

But at 1.57 s, the ball is returning to the balcony’s height, not landing on the ground.

The wall is reached later, at 2.18 s.

Question 3

A ball is launched from a platform 5.0 m above ground level at 18.0 m/s, at \(30^\circ\) above the horizontal.

A student reports two results:

• The ball’s speed at maximum height is zero.
• Its impact speed when it reaches the ground is 18.0 m/s.

Assess both claims. Calculate the correct speed at maximum height and immediately before impact.

Solution 3

Neither reported speed is correct. The speed at maximum height is 15.6 m/s, while the impact speed is approximately 20.5 m/s.

The two errors involve different physical ideas.

Step 1: Resolve the initial velocity.

\[ u_x=18.0\cos30^\circ =15.6\text{ m s}^{-1} \]

\[ u_y=18.0\sin30^\circ =9.0\text{ m s}^{-1} \]

Step 2: Calculate the speed at maximum height.

At maximum height,

\[ v_y=0 \]

However, horizontal velocity remains constant:

\[ v_x=u_x=15.6\text{ m s}^{-1} \]

Therefore, the total speed is

\[ v=\sqrt{v_x^2+v_y^2} \]

\[ v=\sqrt{15.6^2+0^2} \]

\[ \boxed{v=15.6\text{ m s}^{-1}} \]

The student has confused zero vertical velocity with zero total velocity.

Step 3: Calculate the vertical velocity at impact.

The ball lands 5.0 m below its launch point.

Therefore,

\[ \Delta y=-5.0\text{ m} \]

Use the vertical SUVAT equation:

\[ v_y^2=u_y^2+2a_y\Delta y \]

Substitute:

\[ v_y^2=9.0^2+2(-9.8)(-5.0) \]

\[ v_y^2=81+98 \]

\[ v_y^2=179 \]

\[ v_y=\pm\sqrt{179} \]

The ball is descending immediately before impact, so choose the negative value:

\[ v_y=-13.4\text{ m s}^{-1} \]

Step 4: Recombine the velocity components.

Horizontal velocity remains

\[ v_x=15.6\text{ m s}^{-1} \]

The impact speed is

\[ v=\sqrt{v_x^2+v_y^2} \]

\[ v=\sqrt{15.6^2+13.4^2} \]

\[ \boxed{v=20.5\text{ m s}^{-1}} \]

The second student claim is also incorrect.

The ball finishes 5.0 m lower than its launch position. Under the ideal projectile model, gravity increases its kinetic energy as it loses gravitational potential energy.

Its impact speed must therefore be greater than its launch speed.

This gives you another useful consistency check: a projectile landing below its launch height should be moving faster than it was initially, provided air resistance is negligible.

Question 4

A projectile is launched from level ground at 20.0 m/s, at \(45^\circ\) above the horizontal.

A vertical wall is located 30.0 m from the launch point. The wall contains a small opening centred 8.0 m above ground level. The opening is 0.20 m high, extending from 7.90 m to 8.10 m.

A student solves the vertical displacement equation for a height of 8.0 m and obtains two positive times. They use the earlier time and conclude that the projectile misses the opening.

Is this conclusion correct? Determine whether the projectile passes through the opening and explain which vertical crossing is relevant.

Solution 4

The projectile passes through the opening while descending. The earlier vertical crossing is irrelevant because it occurs much closer to the launch point.

Step 1: Resolve the initial velocity.

\[ u_x=20.0\cos45^\circ =14.14\text{ m s}^{-1} \]

\[ u_y=20.0\sin45^\circ =14.14\text{ m s}^{-1} \]

Step 2: Find when the projectile reaches the wall.

Use horizontal motion:

\[ x=u_xt \]

\[ t=\frac{30.0}{14.14} \]

\[ \boxed{t=2.12\text{ s}} \]

Step 3: Calculate the vertical position at the wall.

\[ \Delta y=u_yt-\frac12gt^2 \]

Substitute:

\[ \Delta y=14.14(2.12)-4.9(2.12)^2 \]

\[ \Delta y=30.0-22.05 \]

\[ \boxed{\Delta y=7.95\text{ m}} \]

The opening extends from 7.90 m to 8.10 m.

Since

\[ 7.90<7.95<8.10 \]

the projectile passes through the opening.

Step 4: Explain the two possible vertical crossing times.

For a vertical displacement of 8.0 m:

\[ 8.0=u_yt-\frac12gt^2 \]

Rearrange:

\[ 4.9t^2-14.14t+8.0=0 \]

Using the quadratic formula:

\[ t=\frac{14.14\pm\sqrt{14.14^2-4(4.9)(8.0)}}{9.8} \]

\[ t=\frac{14.14\pm\sqrt{43.2}}{9.8} \]

Therefore,

\[ t_1=0.77\text{ s} \]

\[ t_2=2.11\text{ s} \]

The projectile reaches 8.0 m twice:

• Once while travelling upwards.
• Once while travelling downwards.

The first crossing occurs at a horizontal position of approximately

\[ x_1=14.14(0.77)=10.9\text{ m} \]

The second occurs at approximately

\[ x_2=14.14(2.11)=29.9\text{ m} \]

The wall is at 30.0 m, so the second crossing is the relevant one.

The trap is assuming that a vertical position corresponds to only one time.

A projectile can reach the same height twice. Horizontal position determines which crossing is relevant.

Question 5

A ball is launched from level ground. A high-speed camera records its motion using a calibrated grid and a single shared clock.

The following illustrative data are obtained. Assume the measurements are exact for this question.

A small rectangular opening is positioned ahead of the ball’s path.

Its centre is at \(x=1.60\text{ m}\), \(y=1.40\text{ m}\). It has a horizontal width of 0.20 m and a vertical height of 0.20 m.

Therefore, the opening occupies:

\[ 1.50\leq x\leq1.70 \]

\[ 1.30\leq y\leq1.50 \]

A student claims that the ball can be made to pass through the opening by adjusting its launch angle while keeping the observed motion consistent with the camera data.

Determine whether this is possible.

Your answer must:

• Reconstruct the initial horizontal and vertical velocity components.
• Check whether the measurements are consistent with constant gravitational acceleration.
• Determine whether the projectile passes through any part of the opening, not just its centre.
• Explain whether changing the launch angle can preserve the recorded observations.

Solution 5

The ball cannot pass through the opening, and changing its launch angle would contradict the recorded observations. The measurements uniquely determine both initial velocity components, fixing the trajectory under the stated assumptions.

This question is harder because you must reconstruct the projectile’s motion before testing the target. You cannot simply choose a convenient launch angle.

Step 1: Reconstruct the horizontal velocity.

The horizontal velocity is constant, so

\[ u_x=\frac{x}{t} \]

Using the first measurement:

\[ u_x=\frac{0.40}{0.10} \]

\[ \boxed{u_x=4.0\text{ m s}^{-1}} \]

The second and third measurements give the same result:

\[ \frac{0.80}{0.20}=4.0 \]

\[ \frac{1.20}{0.30}=4.0 \]

The horizontal data are consistent with constant horizontal velocity.

Step 2: Reconstruct the initial vertical velocity.

Use

\[ \Delta y=u_yt-\frac12gt^2 \]

Rearrange:

\[ u_y=\frac{\Delta y+\frac12gt^2}{t} \]

Using the first measurement:

\[ u_y=\frac{0.551+4.9(0.10)^2}{0.10} \]

\[ u_y=\frac{0.551+0.049}{0.10} \]

\[ u_y=6.0\text{ m s}^{-1} \]

Check the second measurement:

\[ u_y=\frac{1.004+4.9(0.20)^2}{0.20} \]

\[ u_y=\frac{1.004+0.196}{0.20} \]

\[ u_y=6.0\text{ m s}^{-1} \]

The third measurement also gives

\[ u_y=\frac{1.359+4.9(0.30)^2}{0.30} \]

\[ u_y=\frac{1.359+0.441}{0.30} \]

\[ u_y=6.0\text{ m s}^{-1} \]

All three measurements agree.

The reconstructed initial velocity is therefore

\[ \boxed{u_x=4.0\text{ m s}^{-1}} \]

\[ \boxed{u_y=6.0\text{ m s}^{-1}} \]

Step 3: Check the launch speed and angle.

The launch speed is

\[ u=\sqrt{u_x^2+u_y^2} \]

\[ u=\sqrt{4.0^2+6.0^2} \]

\[ \boxed{u=7.21\text{ m s}^{-1}} \]

The launch angle is

\[ \theta=\tan^{-1}\left(\frac{u_y}{u_x}\right) \]

\[ \theta=\tan^{-1}\left(\frac{6.0}{4.0}\right) \]

\[ \boxed{\theta=56.3^\circ} \]

The camera data have fixed both the launch speed and launch angle.

Step 4: Calculate the projectile’s position at the centre of the opening.

The opening is centred at

\[ x=1.60\text{ m} \]

Using horizontal motion:

\[ x=u_xt \]

\[ t=\frac{1.60}{4.0} \]

\[ t=0.40\text{ s} \]

Now calculate vertical displacement:

\[ \Delta y=u_yt-\frac12gt^2 \]

\[ \Delta y=6.0(0.40)-4.9(0.40)^2 \]

\[ \Delta y=2.40-0.784 \]

\[ \boxed{\Delta y=1.616\text{ m}} \]

The centre of the opening is at 1.40 m, and its upper boundary is at 1.50 m.

At the centre of the opening’s horizontal position, the ball is therefore above the opening.

But we cannot stop here.

The opening has a width of 0.20 m, so we must check whether the trajectory enters it anywhere between \(x=1.50\) m and \(x=1.70\) m.

Step 5: Test the entire horizontal width of the opening.

At the left edge,

\[ t=\frac{1.50}{4.0}=0.375\text{ s} \]

The vertical displacement is

\[ \Delta y=6.0(0.375)-4.9(0.375)^2 \]

\[ \Delta y=2.25-0.689 \]

\[ \boxed{\Delta y=1.561\text{ m}} \]

This is already above the opening’s upper boundary of 1.50 m.

At the right edge,

\[ t=\frac{1.70}{4.0}=0.425\text{ s} \]

\[ \Delta y=6.0(0.425)-4.9(0.425)^2 \]

\[ \boxed{\Delta y=1.665\text{ m}} \]

The projectile is still travelling upwards throughout this interval because

\[ v_y=u_y-gt \]

At the left edge,

\[ v_y=6.0-9.8(0.375) \]

\[ v_y=2.33\text{ m s}^{-1} \]

Its vertical velocity remains positive across the entire opening.

Therefore, its height increases from approximately 1.561 m to 1.665 m while travelling across the opening’s horizontal width.

The lowest height it reaches within that interval is 1.561 m, which is still above the opening’s upper boundary of 1.50 m.

Step 6: Decide whether changing the launch angle can solve the problem.

No.

The measured horizontal positions determine the horizontal velocity:

\[ u_x=4.0\text{ m s}^{-1} \]

The measured vertical positions, together with the known gravitational acceleration, determine the vertical velocity:

\[ u_y=6.0\text{ m s}^{-1} \]

These components uniquely determine the launch speed, launch angle, and subsequent trajectory.

Changing the angle would change at least one of the initial velocity components. The new projectile would no longer reproduce all three recorded observations.

The tempting approach is to treat the opening as a fresh projectile-launch problem and choose an angle that sends a ball through its centre.

But that would describe a different launch, not the ball recorded by the camera.

This is the key constraint: a proposed answer must satisfy every piece of information about the same projectile, not just the final target position.

06What to check before accepting a projectile answer

When you finish an HSC projectile calculation, run through these checks before moving on.

1. Have you resolved the initial velocity into horizontal and vertical components?
2. Have you used the correct vertical displacement, measured from the chosen origin?
3. Have you used the same time in both horizontal and vertical equations?
4. Have you checked whether the projectile lands at its original height or at a different height?
5. Is the calculated time positive and physically meaningful?
6. Is the calculated maximum height consistent with the target’s position?
7. Does the horizontal distance agree with the horizontal velocity and flight time?
8. Have you interpreted negative vertical velocity as downward motion rather than treating it as a negative speed?

The most important habit is to stop treating projectile formulas as interchangeable shortcuts.

Start with the two component equations, identify the actual endpoint, and make both directions agree at the same time.

Once the trajectory passes these checks, you can move on to calculating its impact velocity. The next useful step is Final Velocity of a Projectile: HSC Physics Guide, where you recombine the horizontal and vertical velocity components to find the projectile’s final speed and direction.