To find the horizontal speed vxv_xvx​ that the ball on the left must start with so that it hits the ground at the same horizontal position as the ball on the right, given that:

• The two balls are dropped from the same height.
• The ball on the right falls straight down with zero initial velocity.
• The ball on the left is given an initial horizontal velocity vxv_xvx​.
• The horizontal distance between the balls at release is 3.0 m.

Step 1: Determine the time ttt it takes for the balls to hit the ground

Since both balls start from the same height hhh and fall freely under gravity (acceleration g≈9.8 m/s2g\approx 9.8,m/s^2g≈9.8m/s2), the time to fall is the same for both and depends only on the vertical motion:

t=2hgt=\sqrt{\frac{2h}{g}}t=g2h​​

This time ttt is the duration both balls take to reach the ground.

Step 2: Relate horizontal displacement of the left ball to the initial

velocity

The ball on the left moves horizontally with constant velocity vxv_xvx​ (no horizontal acceleration), so its horizontal displacement after time ttt is:

x=vx×tx=v_x\times tx=vx​×t

Step 3: Use the initial horizontal separation to find vxv_xvx​

The ball on the left must cover the initial 3.0 m horizontal gap to land at the same position as the ball on the right. Therefore,

vx×t=3.0 mv_x\times t=3.0,mvx​×t=3.0m

Solving for vxv_xvx​:

vx=3.0tv_x=\frac{3.0}{t}vx​=t3.0​

Summary

• Calculate the fall time t=2hgt=\sqrt{\frac{2h}{g}}t=g2h​​.
• Then calculate vx=3.0tv_x=\frac{3.0}{t}vx​=t3.0​.

This vxv_xvx​ is the horizontal speed the left ball must have to land at the same spot as the right ball, despite starting 3.0 m apart horizontally. Example: If the height hhh is 5.0 m, then

t=2×5.09.8≈1.01 st=\sqrt{\frac{2\times 5.0}{9.8}}\approx 1.01,st=9.82×5.0​​≈1.01s

vx=3.01.01≈2.97 m/sv_x=\frac{3.0}{1.01}\approx 2.97,m/svx​=1.013.0​≈2.97m/s

So the ball on the left must start with about 3.0 m/s horizontally to hit the ground at the same position as the ball on the right