A ball is thrown vertically upwards with a veloctty of 20 mathrm{m} mathrm{s}^{-1} from top of a mul

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2.Motion in Straight Line

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A ball is thrown vertically upwards with a veloctty of $20 \;\mathrm{m} \mathrm{s}^{-1}$ from the top of a multistorey building. The helght of the point from where the ball is thrown is $25.0\; \mathrm{m}$ from the ground. how long will it be in $seconds$ before the ball hits the ground? Take $g=10 \;\mathrm{m} \mathrm{s}^{-2}$

$3$

$2$

$5$

$7$

Solution

Let us take the $y$ -axis in the vertically upward direction with zero at the ground

$\text { Now } v_{o} =+20 \mathrm{m} \mathrm{s}^{-1}$

$a =-g=-10 \mathrm{m} \mathrm{s}^{-2}$

$v =0 \mathrm{m} \mathrm{s}^{-1}$

If the ball rises to height $y$ from the point of launch, then using the equation

$v^{2}=v_{0}^{2}+2 a\left(y-y_{0}\right)$

we get

$0=(20)^{2}+2(-10)\left(y-y_{0}\right)$

Solving, we get, $\left(y-y_{0}\right)=20 \mathrm{m}$

we split the path in two parts: the upward motion (A to B) and the downward motion (B to C) and calculate the corresponding time taken $t_{1}$ and $t_{2}$. Since the velocity at $B$ is zero, we have

$v=v_{0}+a t$

$0=20-10 t_{1}$

Or, $\quad t_{1}=2 \mathrm{s}$

This is the time in going from $A$ to $B$. From $B$, or the point of the maximum height, the ball falls freely under the acceleration due to gravity. The ball is moving in negative $y$ direction. We use equation

$y=y_{0}+v_{0} t+\frac{1}{2} a t^{2}$

We have, $y_{0}=45 \mathrm{m}, y=0, v_{0}=0, a=-g=-10 \mathrm{ms}^{-2}$

$0=45+(1 / 2)(-10) t_{2}^{2}$

Solving, we get $t_{2}=3 s$

Therefore, the total time taken by the ball before it hits the ground $=t_{1}+t_{2}=2 s+3 s=5 s$

Standard 11

Physics