Question

A ball is thrown upwards with a velocity of $2m{s}^{-1}$ from a height of $5m$ from the ground. What will be the speed of the ball when it touches the ground? Take the acceleration due to gravity to be $10m{s}^{-2}$.

• A. $\sqrt{96}m{s}^{-1}$
• B. $\sqrt{52}m{s}^{-1}$
• C. $\sqrt{102}m{s}^{-1}$
• D. $\sqrt{104}m{s}^{-1}$

Answer 1

The correct option is D $\sqrt{104}m{s}^{-1}$

Given:
Initial upward velocity, u = $2m{s}^{-1}$
Initial height of the ball, h = $5m$
Final height of the ball, h' = $0m$ (as it touches ground)
Displacement of the ball, s = $-5m$
Acceleration due to gravity, a = $-10m{s}^{-2}$
According to the third equation of motion:
$v^2$ = $u^2$ + $2\left(a\right)\left(s\right)$
$v^2$ = $2^2$ + $2(-10)(-5)$
$v^2=104$
$v$ = $\sqrt{104}m{s}^{-1}$