Question

A ball is thrown vertically upwards with a velocity of $20{ms}^{-1}$ from the top of a multistory building. The height of the point from where the ball is thrown is $025.0m$ from the ground. $\left(a\right)$ How high will the ball rise? And $\left(b\right)$ how long will it be before the ball hits the ground?
Take $g=10m{s}^{-2}$.

Answer 1

a)

From the equation of motion

H = $\frac{u^2}{\left(2g\right)}$

H = $\frac{{20}^2}{(2\times 10)}$

H = 20 m

The ball will reach 20 m high from the point of projection.

It reaches 20 + 20.5 = 40.5m high from the ground.

(b)

$t_1=\frac{2u}{g}$

$t_1=\frac{2\times 20}{10}$

$t_1$= 4 seconds

If v is the velocity with which it hits the ground then,

v $=\sqrt{(u^2+2aS)}$

v = $\sqrt{({20}^2+2\times 10\times 25)}$

v = 30 $m/s$

v = u + at

$t_2=\frac{(v-u)}{g}$

$t_2=\frac{(30-20)}{10}$

$t_2$ = 1 second

Time taken for it to hit the ground = $t_1+t_2$

= 4 + 1

= 5 seconds