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
