Question

A ball is released out a window that is 43 m above the ground. At what speed will it hit the ground? Assume air resistance is negligible.

58 m/s

29 m/s

0.86 km/s

0 m/s

Answer 1

The final speed of the ball is 29 m/s

Explanation:

The motion of the ball is a free fall motion, which is affected by the force of gravity only. Therefore, the ball accelerates towards the ground with a constant acceleration, $g=9.8 m/s^2$ towards the ground (acceleration of gravity). So, we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration

s is the distance covered

For the ball in this problem,

u = 0 (the ball is dropped from rest)

s = 43 m

Solving for v, we find the final velocity:

$v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(43)}=29.0 m/s$

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