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

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Solution

a)
From the equation of motion
H = $\frac{u^{2}}{(2 g)}$

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)
$t_{1} = \frac{2 u}{g}$

$t_{1} = \frac{2 \times 20}{10}$

$t_{1}$ = 4 seconds

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

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

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

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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 025.0 m from the ground. (a) How high will the ball rise? And (b) how long will it be before the ball hits the ground?Take g=10 m s−2.