How to Find a Projectile’s Velocity Angle Before Landing

A ball is kicked from a raised sports field at \(30^\circ\) above the horizontal. It rises, curves back down, and lands on a field 5 metres lower. Will it hit the ground at \(30^\circ\) below the horizontal?

Not quite. That matching angle only works when the ball lands at the same height from which it was launched. With a lower landing point, gravity has more distance to increase its downward speed, so the impact angle is steeper.

To find the angle of a projectile’s velocity just before landing, calculate its final horizontal velocity \(v_x\) and vertical velocity \(v_y\), then use:

\[ \theta=\tan^{-1}\left(\frac{|v_y|}{|v_x|}\right) \]

Here, \(\theta\) is the acute angle between the velocity and the horizontal. For a projectile travelling right and down, report the result as degrees below the horizontal.

The important part is finding the correct components at the instant before impact. The launch angle and landing angle are not automatically equal.

01The impact angle follows the velocity, not the whole flight path

Picture a ball travelling through the air. At any instant, its velocity points in the direction it is moving.

If you could freeze the ball immediately before landing, its velocity would point along the tangent to its curved path at that position.

It would not necessarily point along the straight line connecting the launch point and landing point.

Why? That straight line describes the ball’s overall displacement. We need its instantaneous direction of motion.

To find that direction, split the final velocity into two perpendicular components:

• \(v_x\): horizontal velocity, measured in \(\mathrm{m\,s^{-1}}\).
• \(v_y\): vertical velocity, measured in \(\mathrm{m\,s^{-1}}\).

Choose right as positive \(x\) and up as positive \(y\). A ball moving right and down has \(v_x>0\) and \(v_y<0\).

The actual velocity is the diagonal of this triangle.

Because the horizontal and vertical components are perpendicular, trigonometry gives:

\[ \tan\theta=\frac{|v_y|}{|v_x|} \]

Therefore,

\[ \theta=\tan^{-1}\left(\frac{|v_y|}{|v_x|}\right) \]

The absolute values give the angle’s size. The signs tell us the direction.

For example, a ball with \(v_x=12.0\ \mathrm{m\,s^{-1}}\) and \(v_y=-9.0\ \mathrm{m\,s^{-1}}\) is travelling right and downward. Its direction is:

\[ \begin{aligned} \theta&=\tan^{-1}\left(\frac{9.0}{12.0}\right)\\ &=36.9^\circ \end{aligned} \]

So its velocity is directed \(36.9^\circ\) below the horizontal, towards the right.

This method works regardless of whether the ball was initially launched upwards, horizontally, or downwards, provided we know the velocity components at the relevant instant.

If \(v_x=0\), the projectile is travelling vertically, and you should state its direction directly rather than divide by zero.

02How could a camera measure the landing angle?

Imagine recording a ball’s flight using a stationary camera positioned side-on.

Place a marked vertical measuring scale beside the landing area, in the same plane as the ball’s motion. The ball travels horizontally across the camera’s view, while the scale helps measure its changing height.

Suppose you first measure the straight line between where the ball was released and where it landed.

Would the angle of that line give you the impact angle?

No. It gives the direction of the total displacement, not the direction the ball is moving at the end.

A better method is to track the ball’s position in two frames immediately before impact.

Consider these hypothetical video measurements. Model the ball as a point, with \(y=0\) at its landing height.

The ball travels \(0.22\ \mathrm{m}\) horizontally in \(0.020\ \mathrm{s}\).

Its average horizontal velocity is:

\[ \begin{aligned} v_x&\approx\frac{0.22}{0.020}\\ &=11.0\ \mathrm{m\,s^{-1}} \end{aligned} \]

Its average vertical velocity over those frames is:

\[ \begin{aligned} \bar v_y&=\frac{0.01-0.26}{0.020}\\ &=-12.5\ \mathrm{m\,s^{-1}} \end{aligned} \]

But there is a small limitation. This vertical value describes the average over the interval, not exactly the velocity at its end.

Assuming constant downward acceleration of \(9.8\ \mathrm{m\,s^{-2}}\), the vertical velocity at the second frame is approximately:

\[ \begin{aligned} v_{y,2}&=-12.5-\frac{9.8(0.020)}{2}\\ &=-12.6\ \mathrm{m\,s^{-1}} \end{aligned} \]

The estimated direction near landing is therefore:

\[ \begin{aligned} \theta&\approx\tan^{-1}\left(\frac{12.6}{11.0}\right)\\ &\approx49^\circ \end{aligned} \]

So the ball is approaching the ground at roughly \(49^\circ\) below the horizontal.

These are illustrative readings, not results from a real experiment. A practical measurement would be affected by frame rate, position uncertainty, camera alignment, and potentially air resistance.

The revised two-frame method gets much closer to the instantaneous direction. Using several closely spaced frames and a fitted trajectory would improve the estimate further.

In an HSC calculation, we can usually determine that direction mathematically without a camera.

03Find the horizontal and vertical landing velocities separately

Under the standard ideal projectile model, assume:

• Air resistance is negligible.
• Gravity is constant at \(g=9.8\ \mathrm{m\,s^{-2}}\), directed downward.
• The projectile is treated as a point.
• Right is positive horizontally, and up is positive vertically.
• We calculate the velocity immediately before contact, not after a bounce.

Let \(u\) be the launch speed in \(\mathrm{m\,s^{-1}}\), and \(\alpha\) be the launch angle above the rightward horizontal.

The initial velocity components are:

\[ u_x=u\cos\alpha \]

\[ u_y=u\sin\alpha \]

Here, \(u_x\) and \(u_y\) are the initial horizontal and vertical velocities.

Horizontal motion

Gravity acts vertically. With negligible air resistance, there is no horizontal acceleration.

Therefore:

\[ v_x=u_x=u\cos\alpha \]

The horizontal component stays constant throughout the flight.

Vertical motion

Gravity changes the vertical component every second.

Taking upwards as positive:

\[ v_y=u_y-gt \]

Here, \(t\) is the elapsed time in seconds.

The vertical component can start positive, become zero at maximum height, and then turn negative as the ball descends.

This gives us the most important distinction:

The horizontal velocity is constant, but the vertical velocity is changing.

If you know the flight time, finding the landing components is straightforward. But what if the question doesn’t give you the time?

Choose the vertical equation from the information given

Define the signed vertical displacement:

\[ \Delta y=y_{\mathrm{land}}-y_{\mathrm{launch}} \]

It is zero for a same-height landing, negative when the landing point is lower, and positive when it is higher.

The equation without time,

\[ v_y^2=u_y^2-2g\Delta y \]

is especially useful for uneven landings.

It comes from \(v_y^2=u_y^2+2a_y\Delta y\), using \(a_y=-g\).

Be careful: squaring removes the direction. If the projectile is descending at impact, choose the negative value of \(v_y\).

If the landing surface is at a different height, the same-height shortcut \(v_y=-u_y\) is generally invalid.

For more detail on obtaining the actual contact time, see Projectile Time of Flight for Level and Uneven Landings.

06Questions and solutions

Question 1

A ball is moving horizontally to the right at \(8.0\ \mathrm{m\,s^{-1}}\) and vertically downwards at \(6.0\ \mathrm{m\,s^{-1}}\) immediately before hitting the ground.

Calculate the angle of its velocity relative to the horizontal, and state its direction.

Solution 1

The ball is travelling at \(36.9^\circ\) below the horizontal, towards the right.

With right positive and up positive:

\[ \begin{aligned} v_x&=8.0\ \mathrm{m\,s^{-1}}\\ v_y&=-6.0\ \mathrm{m\,s^{-1}} \end{aligned} \]

Therefore:

\[ \begin{aligned} \theta&=\tan^{-1}\left(\frac{|v_y|}{|v_x|}\right)\\ &=\tan^{-1}\left(\frac{6.0}{8.0}\right)\\ &=36.9^\circ \end{aligned} \]

The downward component makes the angle below horizontal, not above it.

A common mistake is to report \(36.9^\circ\) without specifying the direction. In a velocity question, the direction is part of the answer.

Question 2

A ball is launched from ground level at \(20.0\ \mathrm{m\,s^{-1}}\), making an angle of \(30.0^\circ\) above the horizontal.

A building begins \(20.0\ \mathrm{m}\) horizontally from the launch point. Its horizontal roof is \(4.0\ \mathrm{m}\) above launch height and extends from \(x=20.0\ \mathrm{m}\) to \(x=35.0\ \mathrm{m}\).

Assume the building has a vertical front wall up to roof level, with no overhang.

Determine whether the ball lands on the roof. If it does, calculate its impact angle below the horizontal.

Ignore air resistance, take \(g=9.8\ \mathrm{m\,s^{-2}}\), and treat the ball as a point.

Solution 2

The ball clears the building’s front wall and lands on the roof at \(x=25.9\ \mathrm{m}\), with a velocity directed \(15.0^\circ\) below the horizontal.

The difficult part is deciding which crossing of the roof’s height corresponds to an actual landing.

First resolve the launch velocity:

\[ \begin{aligned} u_x&=20.0\cos30.0^\circ\\ &=17.32\ \mathrm{m\,s^{-1}}\\ u_y&=20.0\sin30.0^\circ\\ &=10.0\ \mathrm{m\,s^{-1}} \end{aligned} \]

At roof height, the vertical displacement is:

\[ \Delta y=+4.0\ \mathrm{m} \]

Use:

\[ v_y^2=u_y^2-2g\Delta y \]

\[ \begin{aligned} v_y^2&=(10.0)^2-2(9.8)(4.0)\\ &=21.6\ \mathrm{m^2\,s^{-2}} \end{aligned} \]

So:

\[ v_y=\pm4.648\ \mathrm{m\,s^{-1}} \]

There are two crossings of this height: one while the ball rises, and another while it falls.

Using \(v_y=u_y-gt\), the ascending crossing occurs at:

\[ \begin{aligned} t_1&=\frac{10.0-4.648}{9.8}\\ &=0.546\ \mathrm{s} \end{aligned} \]

Its horizontal position is:

\[ \begin{aligned} x_1&=u_xt_1\\ &=17.32(0.546)\\ &=9.46\ \mathrm{m} \end{aligned} \]

This is before the building begins, so there is no roof there to land on.

The descending crossing occurs at:

\[ \begin{aligned} t_2&=\frac{10.0+4.648}{9.8}\\ &=1.495\ \mathrm{s} \end{aligned} \]

The corresponding horizontal position is:

\[ \begin{aligned} x_2&=17.32(1.495)\\ &=25.9\ \mathrm{m} \end{aligned} \]

This lies within the roof’s \(20.0\) to \(35.0\ \mathrm{m}\) extent.

We must also check that the ball clears the front wall. At \(x=20.0\ \mathrm{m}\):

\[ t=\frac{20.0}{17.32}=1.155\ \mathrm{s} \]

Its height is:

\[ \begin{aligned} y&=u_yt-\frac12gt^2\\ &=10.0(1.155)\\ &\quad-4.9(1.155)^2\\ &\approx5.01\ \mathrm{m} \end{aligned} \]

The ball is above the \(4.0\ \mathrm{m}\) roof edge, so it clears the wall.

At roof contact, the vertical velocity is negative:

\[ v_y=-4.648\ \mathrm{m\,s^{-1}} \]

Therefore:

\[ \begin{aligned} \theta&=\tan^{-1}\left(\frac{4.648}{17.32}\right)\\ &=15.0^\circ \end{aligned} \]

The tempting mistake is to calculate the angle at the first crossing of \(y=4.0\ \mathrm{m}\). That crossing is real, but the ball is nowhere near the roof.

A height alone does not identify the landing event. The horizontal position must also match the physical landing surface.

Question 3

A ball rolls horizontally off a ledge and lands on flat ground below.

An illustrative slow-motion recording shows that its velocity immediately before impact makes an angle of \(60.0^\circ\) below the horizontal. The landing point is \(18.0\ \mathrm{m}\) horizontally from the ledge.

Find:

• The vertical height of the ledge.
• The ball’s initial horizontal speed.
• The new landing distance and impact angle if the ball leaves the same ledge at twice its original horizontal speed.

Assume the ledge extends behind the launch point, the lower ground extends far enough for both landings, \(g=9.8\ \mathrm{m\,s^{-2}}\), and air resistance is negligible.

Solution 3

The ledge is \(15.6\ \mathrm{m}\) high, and the original horizontal speed is \(10.1\ \mathrm{m\,s^{-1}}\). Doubling that speed increases the landing distance to \(36.0\ \mathrm{m}\) and reduces the impact angle to \(40.9^\circ\) below the horizontal.

The key is that the ball is launched horizontally, so its initial vertical velocity is zero.

Let \(H\) be the ledge height, \(x=18.0\ \mathrm{m}\) the horizontal displacement, and \(t\) the flight time.

Horizontal motion gives:

\[ x=v_xt \]

Vertical motion gives:

\[ |v_y|=gt \]

At impact:

\[ \tan60.0^\circ=\frac{gt}{v_x} \]

Since \(v_x=x/t\):

\[ \tan60.0^\circ=\frac{gt^2}{x} \]

But the vertical drop is:

\[ H=\frac12gt^2 \]

Therefore:

\[ H=\frac{x\tan60.0^\circ}{2} \]

Substitute:

\[ \begin{aligned} H&=\frac{18.0\tan60.0^\circ}{2}\\ &=15.6\ \mathrm{m} \end{aligned} \]

Now calculate the flight time:

\[ \begin{aligned} t&=\sqrt{\frac{2H}{g}}\\ &=\sqrt{\frac{2(15.588)}{9.8}}\\ &=1.78\ \mathrm{s} \end{aligned} \]

The initial horizontal speed is:

\[ \begin{aligned} u_x&=\frac{x}{t}\\ &=\frac{18.0}{1.7836}\\ &=10.1\ \mathrm{m\,s^{-1}} \end{aligned} \]

Now double the launch speed.

The vertical motion is unchanged because the launch height and initial vertical velocity are unchanged. Both balls spend the same \(1.78\ \mathrm{s}\) in the air.

Thus the second ball travels:

\[ \begin{aligned} x_2&=2u_xt\\ &=2(18.0)\\ &=36.0\ \mathrm{m} \end{aligned} \]

The downward impact speed of each ball is:

\[ \begin{aligned} |v_y|&=gt\\ &=9.8(1.7836)\\ &=17.48\ \mathrm{m\,s^{-1}} \end{aligned} \]

The second ball’s horizontal speed is approximately \(20.18\ \mathrm{m\,s^{-1}}\).

Its impact angle becomes:

\[ \begin{aligned} \theta_2&=\tan^{-1}\left(\frac{17.48}{20.18}\right)\\ &=40.9^\circ \end{aligned} \]

A tempting mistake is to calculate the height as \(18.0\tan60^\circ=31.2\ \mathrm{m}\).

That treats the impact angle as the angle of the straight displacement from launch to landing.

For a horizontal launch under constant gravity, the slope of the final velocity is twice the slope of that total displacement. That is why the correct height includes the factor of one-half.

Question 4

A ball is kicked from the top of a straight grassy hillside that slopes downwards to the right.

The hill drops \(1.0\ \mathrm{m}\) vertically for every \(4.0\ \mathrm{m}\) horizontally, and continues with this constant slope.

The ball is launched at \(16.0\ \mathrm{m\,s^{-1}}\) at \(30.0^\circ\) above the horizontal.

Calculate:

• Its velocity angle below the horizontal immediately before landing.
• The acute angle between its velocity and the sloping ground at impact.

Assume \(g=9.8\ \mathrm{m\,s^{-2}}\), no air resistance, and a point projectile.

Solution 4

The ball lands at \(47.1^\circ\) below the horizontal. Its velocity makes an acute angle of \(33.1^\circ\) with the downhill surface.

The key difference from a flat-ground problem is that the landing height changes with horizontal position.

Resolve the initial velocity:

\[ \begin{aligned} u_x&=16.0\cos30.0^\circ\\ &=13.86\ \mathrm{m\,s^{-1}}\\ u_y&=16.0\sin30.0^\circ\\ &=8.00\ \mathrm{m\,s^{-1}} \end{aligned} \]

Because the slope drops \(1\ \mathrm{m}\) for every \(4\ \mathrm{m}\) horizontally, the ground height relative to launch is:

\[ y=-\frac{x}{4} \]

The projectile’s horizontal position is:

\[ x=u_xt \]

Its vertical position is:

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

At landing, the projectile and the ground have the same height:

\[ u_yt-\frac12gt^2=-\frac{u_xt}{4} \]

Rearrange:

\[ t\left(u_y+\frac{u_x}{4}-\frac12gt\right)=0 \]

The solution \(t=0\) is the launch instant. For the later intersection:

\[ t=\frac{2(u_y+u_x/4)}{g} \]

Substitute:

\[ \begin{aligned} t&=\frac{2(8.00+13.86/4)}{9.8}\\ &=2.34\ \mathrm{s} \end{aligned} \]

Now find the vertical velocity:

\[ \begin{aligned} v_y&=u_y-gt\\ &=8.00-9.8(2.3396)\\ &=-14.93\ \mathrm{m\,s^{-1}} \end{aligned} \]

The horizontal velocity remains \(13.86\ \mathrm{m\,s^{-1}}\).

Therefore:

\[ \begin{aligned} \theta&=\tan^{-1}\left(\frac{14.93}{13.86}\right)\\ &=47.1^\circ \end{aligned} \]

This is the velocity angle below the true horizontal.

But the hillside is itself tilted downwards. Its angle below horizontal is:

\[ \begin{aligned} \phi&=\tan^{-1}\left(\frac14\right)\\ &=14.0^\circ \end{aligned} \]

The acute angle between the incoming velocity and the ground is:

\[ \begin{aligned} \theta_{\mathrm{surface}}&=47.1^\circ-14.0^\circ\\ &=33.1^\circ \end{aligned} \]

The tempting shortcut is to say that the ball lands at \(30^\circ\) below horizontal because it was launched at \(30^\circ\) above it.

But the hill falls away from the ball, allowing it to spend longer descending before contact.

Horizontal and vertical motion still follow their separate equations. What couples them is the landing condition \(y=-x/4\).

A second possible mistake is to report the velocity angle relative to horizontal when the question asks for the angle relative to the surface. Those are two different reference directions.

Question 5

Two identical balls are launched simultaneously from the edge of a roof, both travelling towards the right with the same unknown initial speed \(u\).

Ball A is launched at \(30.0^\circ\) above the horizontal.

Ball B is launched at \(30.0^\circ\) below the horizontal.

Both balls land on the same flat ground below the roof. The roof extends only behind the launch point, and neither ball encounters any obstacle before landing.

Measurements show that both balls approach the ground at exactly \(45.0^\circ\) below the horizontal. Their landing points are separated by \(24.0\ \mathrm{m}\) horizontally.

Determine:

• The common launch speed \(u\).
• The roof height \(H\).
• The flight time of each ball.
• Which ball lands first, and why both have the same impact angle.

Finally, decide whether their impact angles would still be equal if the common ground level were changed, while keeping the launch speeds and directions the same.

Use \(g=9.8\ \mathrm{m\,s^{-2}}\), and neglect air resistance.

Solution 5

The common launch speed is \(16.5\ \mathrm{m\,s^{-1}}\), the roof is \(6.93\ \mathrm{m}\) high, Ball A lands after \(2.30\ \mathrm{s}\), and Ball B lands after \(0.616\ \mathrm{s}\).

Ball B lands first, but both have the same impact direction. In fact, their impact angles would remain equal for any shared landing height under the ideal model, although those angles would not necessarily remain \(45^\circ\).

The surprising part is that the ball launched upwards spends much longer in the air, yet does not arrive with a greater downward speed.

To see why, start with the components.

Both balls have the same horizontal velocity:

\[ v_x=u\cos30^\circ \]

Their initial vertical velocities are opposite:

\[ \begin{aligned} u_{y,A}&=u\sin30^\circ\\ u_{y,B}&=-u\sin30^\circ \end{aligned} \]

Both land a vertical displacement \(\Delta y=-H\) below their starting point.

Using:

\[ v_y^2=u_y^2-2g\Delta y \]

we obtain, for either ball:

\[ v_y^2=u^2\sin^230^\circ+2gH \]

The initial vertical velocities have opposite signs, but their squares are identical.

Because both balls are descending when they land:

\[ v_y=-\sqrt{u^2\sin^230^\circ+2gH} \]

So both have exactly the same final horizontal and vertical velocities, provided they share the same launch speed magnitude and landing height.

That explains why the impact directions match. It does not yet tell us their common launch speed.

We can use the measured \(45^\circ\) direction.

At \(45^\circ\) below horizontal:

\[ |v_y|=v_x \]

Therefore:

\[ u^2\sin^230^\circ+2gH =u^2\cos^230^\circ \]

Rearrange:

\[ 2gH=u^2(\cos^230^\circ-\sin^230^\circ) \]

Since \(\cos^230^\circ-\sin^230^\circ=0.5\):

\[ H=\frac{u^2}{4g} \]

Now we have a relationship between the unknown height and launch speed. We need the separation of the landing points to determine their values.

The final vertical velocity of both balls is:

\[ v_y=-u\cos30^\circ \]

For Ball A:

\[ v_y=u\sin30^\circ-gt_A \]

So:

\[ t_A=\frac{u(\sin30^\circ+\cos30^\circ)}{g} \]

For Ball B:

\[ v_y=-u\sin30^\circ-gt_B \]

Therefore:

\[ t_B=\frac{u(\cos30^\circ-\sin30^\circ)}{g} \]

Subtract the flight times:

\[ \begin{aligned} t_A-t_B&=\frac{2u\sin30^\circ}{g}\\ &=\frac{u}{g} \end{aligned} \]

Both balls move horizontally at \(u\cos30^\circ\).

Their difference in horizontal landing distance is therefore:

\[ \Delta x=u\cos30^\circ(t_A-t_B) \]

The measured separation is \(24.0\ \mathrm{m}\), so:

\[ 24.0=\frac{u^2\cos30^\circ}{9.8} \]

Rearrange:

\[ \begin{aligned} u^2&=\frac{24.0(9.8)}{\cos30^\circ}\\ &=271.6\ \mathrm{m^2\,s^{-2}}\\ u&=16.5\ \mathrm{m\,s^{-1}} \end{aligned} \]

Now calculate the roof height:

\[ H=\frac{u^2}{4g} \]

\[ \begin{aligned} H&=\frac{271.6}{4(9.8)}\\ &=6.93\ \mathrm{m} \end{aligned} \]

Using the unrounded speed \(u\approx16.48\ \mathrm{m\,s^{-1}}\), calculate the flight times.

For Ball A:

\[ \begin{aligned} t_A&=\frac{16.48(0.5+0.8660)}{9.8}\\ &=2.30\ \mathrm{s} \end{aligned} \]

For Ball B:

\[ \begin{aligned} t_B&=\frac{16.48(0.8660-0.5)}{9.8}\\ &=0.616\ \mathrm{s} \end{aligned} \]

Ball B lands first because it begins with a downward vertical velocity. Ball A initially rises, loses that upward velocity, then falls to the same ground.

As a consistency check, the common horizontal speed is:

\[ v_x=16.48\cos30^\circ=14.27\ \mathrm{m\,s^{-1}} \]

The horizontal landing distances are:

\[ \begin{aligned} x_A&=14.27(2.2971)\\ &=32.8\ \mathrm{m}\\ x_B&=14.27(0.6155)\\ &=8.78\ \mathrm{m} \end{aligned} \]

Their separation is:

\[ 32.8-8.78\approx24.0\ \mathrm{m} \]

Everything is consistent with the observations.

Now consider the final conceptual question.

Would the impact angles remain equal if both balls landed at a different shared height?

Yes. For either ball:

\[ v_y^2=u^2\sin^230^\circ+2gH \]

Changing \(H\) changes the final vertical speed of both balls by the same amount. Since their horizontal speeds also remain equal, their impact angles remain equal.

However, the new angle would generally not be \(45^\circ\).

The trap is assuming that a longer flight must produce a greater downward impact speed.

Ball A spends longer accelerating downward, but it also begins with an upward vertical velocity. Ball B begins with the opposite vertical velocity and reaches the ground sooner.

For equal initial speed magnitudes and equal vertical displacements, those effects produce the same final vertical speed magnitude.

The strongest general principle is this: under constant gravity and negligible air resistance, the vertical velocity at a particular height depends on the initial vertical velocity squared and the vertical displacement, not on whether the initial vertical velocity pointed up or down.

That is why two very different trajectories can arrive with the same velocity direction.

Once you have the impact angle, you can also calculate the complete final velocity by combining the horizontal and vertical components using Pythagoras. The Final Velocity of a Projectile: HSC Physics Guide develops that next step.

For an angle question, though, keep the central decision simple: identify the actual landing point, determine the two velocity components immediately before contact, and use their ratio to find the direction. The geometry of the landing surface decides when the calculation ends; the velocity components tell you which way the projectile is moving when it gets there.