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किसी बहुमंजले भवन की ऊपरी छत से कोई गेंद् $20\, m s ^{-1}$ के वेग से ऊपर की और ऊधर्वाधर दिशा में फेंकी गई है । जिस बिंदु से गेंद फेंकी गई है धरती से उसकी ऊंचाई $25.0\, m$ है । गेंद धरती से टकराने के पहले कितना समय लेगी ? $g=10 m s ^{-2}$
$3$
$2$
$5$
$7$
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 $=t_{1}+t_{2}=2 s+3 s=5 s$
