Why a Dropped Ball and a Horizontally Launched Ball Land Together

Release one ball from the edge of a table while launching another horizontally from exactly the same height. Do both balls hit the floor at the same time, or does the horizontally launched ball take longer because it travels further?

They hit the floor at the same time, provided they are released simultaneously, have zero initial vertical velocity, and air resistance is negligible.

The reason is that horizontal and vertical motion are independent in the ideal projectile model. The launched ball moves sideways while falling, but its horizontal speed does not change how quickly it falls.

This is a useful result in HSC Physics because it lets you solve the vertical and horizontal parts of projectile motion separately.

01Why do both balls land together?

Start by looking at what happens vertically.

Imagine two identical steel balls positioned 1.25 m above a level floor:

• Ball A is simply released from rest.
• Ball B is launched horizontally at 3.0 m/s.

Both balls start at exactly the same height. Neither has any initial vertical velocity.

Now consider what gravity does.

Gravity accelerates both balls vertically downwards at approximately \(9.8\text{ m/s}^2\). It doesn’t matter whether a ball is stationary horizontally or travelling at 3.0 m/s.

After 0.25 seconds, both balls have fallen the same vertical distance. After 0.50 seconds, they are still at the same height.

Ball B has travelled horizontally, but it hasn’t fallen any further or any less than ball A.

This is the key idea: horizontal velocity does not affect vertical acceleration when air resistance is neglected.

Separate the horizontal and vertical motion

To describe the motion mathematically, choose:

• The horizontal direction as the positive \(x\)-axis.
• The upwards direction as the positive \(y\)-axis.
• The floor as \(y=0\).
• The initial height of both balls as \(h\).

We use \(g=9.8\text{ m/s}^2\) for gravitational acceleration.

The horizontal and vertical accelerations are:

\[ a_x=0 \]

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

Notice that gravity only produces vertical acceleration in this model.

The two displacement equations are therefore:

Horizontal motion:

\[ x-x_0=u_xt \]

Vertical motion:

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

Here:

• \(x_0\) is the initial horizontal position, in metres.
• \(u_x\) is the initial horizontal velocity, in m/s.
• \(u_y\) is the initial vertical velocity, in m/s.
• \(t\) is elapsed time, in seconds.
• \(h\) is the initial height, in metres.
• \(g\) is gravitational acceleration, in m/s².

Now compare the two balls.

The important observation is that both balls have identical vertical conditions.

For each ball, the vertical displacement equation becomes:

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

When either ball reaches the floor, \(y=0\). Therefore:

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

Rearranging:

\[ \frac12gt^2=h \]

\[ t^2=\frac{2h}{g} \]

Taking the positive square root:

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

This is the time taken for a ball to fall from height \(h\), starting with zero vertical velocity.

Notice something important: there is no horizontal velocity in this equation.

Both balls have the same \(h\), the same initial vertical velocity, and the same gravitational acceleration. They must therefore have the same flight time.

See how their positions change

For the example above, the initial height is 1.25 m and ball B has a horizontal velocity of 3.0 m/s.

The following values are calculated using the ideal projectile equations.

Both balls reach the floor after approximately 0.505 seconds.

The difference is their horizontal position. Ball B travels approximately 1.52 m before landing, while ball A stays directly below its release point.

The curved trajectory doesn’t mean ball B falls more slowly. It simply combines horizontal movement with the same vertical fall.

02How an experiment tests this prediction

The independence of horizontal and vertical motion can be demonstrated using a simple two-ball experiment.

The question is straightforward: does giving a ball horizontal velocity change the time it takes to fall?

Setting up the apparatus

Use two identical steel balls.

1. Position ball A in a release clamp above a level floor.
2. Position ball B in a horizontal launcher at exactly the same height.
3. Connect the release clamp and launcher to a shared trigger.
4. Arrange a high-speed camera so that it can record both balls from the side.
5. Place clear landing markers beneath the expected paths.

The shared trigger releases ball A and launches ball B simultaneously.

What should happen?

There are two possible predictions.

If horizontal motion affects the rate of falling, ball B should take a different amount of time to reach the floor.

If horizontal and vertical motion are independent, both balls should reach the floor simultaneously.

The second prediction follows from the equations of motion.

A high-speed recording should show both balls reaching the floor in the same frame, or within the recording’s measurement uncertainty.

Interpreting the measurements

Suppose the initial height is 1.25 m.

Our theoretical calculation gives:

\[ t=\sqrt{\frac{2(1.25)}{9.8}} \]

\[ t=0.505\text{ s} \]

A camera recording at 240 frames per second captures approximately:

\[ N=240(0.505) \]

\[ N\approx121\text{ frames} \]

Here is an illustrative set of measurements from such a recording.

These are illustrative measurements, not results from an actual trial.

The difference is approximately one camera frame:

\[ \Delta t=\frac{1}{240}=0.00417\text{ s} \]

A difference this small is consistent with the theoretical prediction, given the recording resolution and uncertainty in identifying the exact release and impact frames.

Why might an experiment initially give different results?

A poorly controlled experiment can make one ball appear to land later.

For example, imagine one student releasing ball A by hand while another operates the horizontal launcher. Even a small delay between the two actions can change the absolute landing times.

The balls might also start at slightly different heights, or the launcher might be tilted upwards or downwards.

A better design uses:

• One shared release mechanism.
• Carefully measured starting heights.
• A launcher checked to ensure it is horizontal.
• Identical balls.
• High-speed video to compare release and landing times.

The experiment illustrates the principle of independent horizontal and vertical motion. It does not establish that all real projectiles behave perfectly in every condition.

03An important distinction: horizontal launch versus same-level launch

There is a common source of confusion when using projectile equations.

The two balls in our experiment start at the same height as each other. However, they do not land at that height. They land on the floor below.

This is a projectile launched from a height, not a projectile that returns to its launch height.

These are different situations.

For a projectile launched upwards and returning to its original height, the vertical displacement equation is:

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

Taking the non-zero solution:

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

This formula assumes a positive initial vertical velocity and that the projectile lands at its original height.

It is not the appropriate formula for a ball launched horizontally from a table.

For a horizontal launch, the initial vertical velocity is zero, but the ball lands below its launch point. You must use the launch-height equation instead.

A useful decision process

Whenever you encounter a projectile question, work through these steps:

1. Identify the initial and final vertical positions.
2. Determine the initial vertical velocity.
3. Use vertical acceleration to calculate the flight time.
4. Substitute that time into the horizontal equation to find horizontal displacement.

This prevents you from accidentally using a same-level projectile formula for a launch from a height.

04Worked examples

05Questions and solutions

Question 1

Two balls are released simultaneously from a height of 0.80 m above a level floor.

Ball A is dropped vertically from rest. Ball B is launched horizontally at 2.5 m/s.

Assume negligible air resistance.

Calculate:

(a) The time taken for each ball to reach the floor.

(b) The horizontal distance travelled by ball B before impact.

Solution 1

Both balls reach the floor after approximately 0.404 seconds, and ball B travels approximately 1.01 m horizontally.

Part (a): Flight time

Both balls have zero initial vertical velocity and experience the same gravitational acceleration.

Use:

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

Substitute:

\[ t=\sqrt{\frac{2(0.80)}{9.8}} \]

\[ t=\sqrt{0.1633} \]

\[ \boxed{t=0.404\text{ s}} \]

The horizontal launch speed does not appear in this equation.

Part (b): Horizontal distance

For ball B:

\[ x-x_0=u_xt \]

Substitute:

\[ x-x_0=2.5(0.404) \]

\[ \boxed{x-x_0=1.01\text{ m}} \]

Ball B travels 1.01 m horizontally before landing.

Both balls take the same time to fall, but their horizontal displacements are different.

Question 2

Two balls are released simultaneously above the same level floor.

• Ball A is dropped from a height of 2.45 m.
• Ball B is launched horizontally at 7.0 m/s from a height of 1.80 m.

Neglect air resistance.

(a) Determine which ball reaches the floor first and calculate the time difference.

(b) What initial vertical velocity would ball B need to have to land at exactly the same time as ball A?

Use upwards as the positive vertical direction.

Solution 2

Ball B lands first, approximately 0.101 seconds before ball A. To make both balls land together, ball B would need an initial upward vertical velocity of approximately 0.92 m/s.

The important difference is that the balls start at different heights.

Part (a): Compare the flight times

For ball A:

\[ t_A=\sqrt{\frac{2(2.45)}{9.8}} \]

\[ \boxed{t_A=0.707\text{ s}} \]

For ball B, its initial vertical velocity is zero because it is launched horizontally.

\[ t_B=\sqrt{\frac{2(1.80)}{9.8}} \]

\[ \boxed{t_B=0.606\text{ s}} \]

Calculate the difference:

\[ \Delta t=0.707-0.606 \]

\[ \boxed{\Delta t=0.101\text{ s}} \]

Ball B reaches the floor first.

Its horizontal speed does not make it fall faster. It simply starts closer to the floor.

Part (b): Find the required vertical velocity

For both balls to land together, ball B must have the same flight time as ball A:

\[ t_B=0.707\text{ s} \]

Use the vertical displacement equation for ball B:

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

At the floor, \(y=0\).

Substitute:

\[ 0=1.80+u_y(0.707) -4.9(0.707)^2 \]

Since:

\[ 4.9(0.707)^2\approx2.45 \]

We obtain:

\[ 0=1.80+0.707u_y-2.45 \]

Rearranging:

\[ 0.707u_y=0.65 \]

\[ u_y=\frac{0.65}{0.707} \]

\[ \boxed{u_y=+0.92\text{ m/s}} \]

Ball B needs an initial upward vertical velocity of approximately 0.92 m/s.

This is an important distinction: changing its horizontal speed cannot make the two balls land together. Changing its initial vertical velocity can.

Question 3

A physics student investigates the simultaneous landing of a dropped ball and a horizontally launched ball using a camera recording at 240 frames per second.

Both balls start at a height of 1.25 m.

The following illustrative frame numbers are recorded:

The balls were released using two separate hand-operated mechanisms.

(a) Calculate the flight time of each ball.

(b) Did the balls hit the floor simultaneously?

(c) Does this experiment contradict the prediction that horizontal and vertical motion are independent?

(d) Suggest two improvements to the experimental design.

Solution 3

Both balls have the same measured flight time of approximately 0.504 seconds. They do not hit the floor simultaneously because ball B was released three frames later.

Part (a): Calculate each flight time

For ball A:

\[ t_A=\frac{121}{240} \]

\[ \boxed{t_A=0.504\text{ s}} \]

For ball B, calculate the number of frames between its own release and impact:

\[ 124-3=121\text{ frames} \]

Therefore:

\[ t_B=\frac{121}{240} \]

\[ \boxed{t_B=0.504\text{ s}} \]

Both balls have identical measured flight times.

Part (b): Did they land simultaneously?

No.

Ball A lands at frame 121, while ball B lands at frame 124.

The difference is:

\[ \Delta t=\frac{124-121}{240} \]

\[ \boxed{\Delta t=0.0125\text{ s}} \]

Ball B lands 0.0125 seconds later in absolute time.

However, it was also released 0.0125 seconds later.

Part (c): Does this contradict the prediction?

No. The results actually support the prediction.

Both balls took the same amount of time to fall after their respective releases.

The difference in landing times is explained by the difference in release times, not by horizontal velocity.

This is why it is important to distinguish flight duration from the time at which an object lands.

Part (d): Improve the experiment

Two useful improvements are:

1. Connect both release mechanisms to a single trigger so that the balls start moving simultaneously.
2. Use a high-speed camera to identify the precise release and impact frames, repeating the experiment several times.

The initial heights and horizontal alignment of the launcher should also be checked.

A controlled experiment must isolate the effect being investigated. Otherwise, timing errors can be mistaken for a physical effect.

Question 4

Two identical steel balls are positioned at the same height above a level floor.

Ball A is dropped vertically, while ball B is launched horizontally at 18 m/s.

A student claims:

“Both balls must land at exactly the same time, even in real air, because gravity accelerates both balls at 9.8 m/s².”

Assume the balls move through still air and that air resistance follows the standard quadratic drag model, in which drag force is proportional to the square of speed and acts opposite to the velocity.

Evaluate the student’s claim.

Explain which ball is expected to land later and why the ideal projectile model does not guarantee equal landing times in this situation.

Solution 4

The student’s claim is incorrect. Under the stated quadratic-drag model, the horizontally launched ball is expected to land later.

The key issue is the difference between gravitational acceleration and total acceleration.

In the ideal projectile model, gravity is the only force acting on each ball after release.

Therefore, both balls have identical vertical acceleration:

\[ a_y=-g \]

Their vertical motions are identical.

However, real air exerts a drag force opposite to the direction of motion.

Ball A moves vertically downwards, so its drag force acts vertically upwards.

Ball B moves both horizontally and vertically. Its velocity has two components, so its drag force acts opposite the combined velocity.

As ball B falls, its drag force has an upward vertical component.

Under quadratic drag, the magnitude of this upward component depends on the total speed. The horizontal component of velocity therefore affects the vertical drag experienced by ball B.

At the same downward vertical speed, ball B has a greater total speed than ball A. Its upward vertical drag component is consequently greater under this model.

This reduces its downward acceleration compared with ball A.

The result is that ball B takes longer to reach the floor.

The important distinction is:

• Gravity still accelerates both balls downwards at \(9.8\text{ m/s}^2\).
• Their net vertical accelerations are no longer necessarily equal because air resistance also acts on them.

An exact difference in landing times cannot be calculated from height and horizontal speed alone. You would also need information such as the balls’ mass and diameter, air density, and drag coefficient.

For standard HSC projectile calculations, air resistance is normally neglected unless the question specifically introduces it.

Question 5

This final problem changes one detail that is easy to overlook: the horizontal launcher is moving vertically.

Ball A is released from rest from a stationary platform 4.90 m above a level floor.

At exactly the same instant, ball B is launched horizontally at 5.0 m/s relative to a lift that is rising vertically at a constant speed of 2.0 m/s.

At the instant of launch, the lift’s launcher is also 4.90 m above the floor.

Assume:

• The lift moves vertically upwards at constant velocity.
• The launcher points horizontally relative to the lift.
• Both balls are released at the same instant.
• Air resistance is negligible.

(a) Will the balls reach the floor simultaneously?

(b) Calculate the flight time of each ball.

(c) Calculate the horizontal distance travelled by ball B before it lands.

Solution 5

No. Ball B lands approximately 0.225 seconds after ball A.

The crucial detail is that ball B inherits the lift’s upward velocity when it leaves the launcher. Although it is launched horizontally relative to the lift, it does not have zero vertical velocity relative to the ground.

This changes its vertical motion.

Part (a): Identify the difference in initial conditions

Ball A is released from a stationary platform.

Therefore:

\[ u_{y,A}=0 \]

Ball B is launched from a lift moving upwards at 2.0 m/s.

The ball inherits this vertical velocity at release.

Therefore, relative to the ground:

\[ u_{y,B}=+2.0\text{ m/s} \]

The positive sign indicates upwards motion.

Both balls start at the same height, but their initial vertical velocities are different.

This means their vertical equations are no longer identical.

Part (b): Calculate both flight times

For ball A:

\[ 0=4.90-\frac12(9.8)t_A^2 \]

Rearranging:

\[ t_A=\sqrt{\frac{2(4.90)}{9.8}} \]

\[ \boxed{t_A=1.00\text{ s}} \]

Now consider ball B.

Its initial vertical velocity is \(+2.0\text{ m/s}\).

Use:

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

Substitute:

\[ 0=4.90+2.0t_B-4.9t_B^2 \]

Rearrange:

\[ 4.9t_B^2-2.0t_B-4.90=0 \]

Using the quadratic formula:

\[ t_B= \frac{2.0+\sqrt{(-2.0)^2+ 4(4.9)(4.90)}}{9.8} \]

\[ \boxed{t_B=1.225\text{ s}} \]

We use the positive root because time after release must be positive.

Compare the results:

Ball B lands approximately:

\[ 1.225-1.000=0.225\text{ s} \]

later than ball A.

Part (c): Calculate the horizontal distance travelled by ball B

The lift has no horizontal velocity, so ball B’s horizontal velocity relative to the ground is still 5.0 m/s.

Horizontal acceleration is zero.

Therefore:

\[ x-x_0=u_xt \]

Substitute:

\[ x-x_0=5.0(1.225) \]

\[ \boxed{x-x_0=6.13\text{ m}} \]

Ball B travels approximately 6.13 m horizontally before reaching the floor.

Why this result is important

The original two-ball experiment requires both balls to have the same initial vertical velocity.

It isn’t enough for them to start at the same height and at the same instant.

Ball B’s upward velocity comes from the motion of the lift. Its horizontal launch direction, measured relative to the lift, does not remove that inherited vertical velocity.

If the lift were stationary, both balls would have zero initial vertical velocity and would land together.

This is a useful reminder when solving projectile questions: always identify the velocity components relative to the ground, not just relative to the object doing the launching.

Once you can calculate the flight time from vertical motion, the next step is combining the horizontal and vertical equations to describe the projectile’s complete path.

For that, see How to Derive the Parabolic Projectile Path Equation.

You can also practise using the horizontal position to determine vertical height with How to Find a Projectile’s Height at a Horizontal Distance.