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Question: A ball is thrown vertically upwards with a velocity of 20 ms<sup>-1</sup> from the top of a multisto...

A ball is thrown vertically upwards with a velocity of 20 ms-1 from the top of a multistory building of 25 m high. The time taken by the ball to reach the ground is

A

2 s

B

3 s

C

5 s

D

7 s

Answer

5 s

Explanation

Solution

Let t1t _ { 1 } be the time taken by the ball to reach the highest point.

Here , v=0,u=20 ms−1,a=−g=−10 ms−2,t=t1\mathrm { v } = 0 , \mathrm { u } = 20 \mathrm {~ms} ^ { - 1 } , \mathrm { a } = - \mathrm { g } = - 10 \mathrm {~ms} ^ { - 2 } , \mathrm { t } = \mathrm { t } _ { 1 }

As v = u +at

∴\therefore0 = 20 + (-10 ) t1t _ { 1 } or t1t _ { 1 } = 2s

Taking vertical downward motion of the ball from highest point to ground.

Here , u = 0, a = + g = 10 ms−210 \mathrm {~ms} ^ { - 2 } s= 20 m + 25m = 45m t=t2\mathrm { t } = \mathrm { t } _ { 2 }

As S=ut+12at2S = u t + \frac { 1 } { 2 } a t ^ { 2 }

∴45=0+12(10)t22\therefore 45 = 0 + \frac { 1 } { 2 } ( 10 ) \mathrm { t } _ { 2 } ^ { 2 } t22=45×210=9010=9\mathrm { t } _ { 2 } ^ { 2 } = \frac { 45 \times 2 } { 10 } = \frac { 90 } { 10 } = 9 \quad or t2=3 s\quad \mathrm { t } _ { 2 } = 3 \mathrm {~s}

Total times taken by the ball to each the ground. =t1+t2=2 s+3 s=5 s= \mathrm { t } _ { 1 } + \mathrm { t } _ { 2 } = 2 \mathrm {~s} + 3 \mathrm {~s} = 5 \mathrm {~s}