Question

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

Answer 1

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$,

where

• $u$ is the initial velocity of the object,
• $v$ is the final velocity of the object.
• $a$ is its acceleration, and
• $x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$. By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$. Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$.

$v^2 - u^2 = 2\, g \cdot h$.

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$.

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$.