Question

A ball is thrown with a velocity of $5m/s$ towards the ground from the top of a building of height $10m$. What is the change in its speed, when the ball is about to hit the ground? Take $g=10m/s^2$.

Answer 1

We have been given:
$u=-5m/s$
$h=-10m$
$a=-10m/s^2$

Using 3rd equation of kinematics:
$v^2=u^2+2as$
$v^2=(-5{)}^2+2(-10\left)\right(-10)$
$v^2=25+200=225$
$v=\sqrt{225}=\pm 15m/s$
As the ball is travelling downwards, the final velocity will also be in downward direction (negative).
Hence, $v=-15m/s$

So, the change in speed of the ball:
$|v|-|u|=15-5=10m/s$