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A man is standing on the top of a $60 m$ high tower. He throws a ball vertically upwards with a velocity of $20 m / s$ . After what time will the ball pass him going downwards? (Take $g = 10 m / s^{2}$ )

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1 s

2 s

3 s

4 s

answer is D.

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Detailed Solution

After

4 s

, the ball will pass him going downwards.
Given data and assumptions:
Initial velocity,

u = 20 m / s

Vertical distance covered,

h = 0 m

(as the ball returns to the man)
Acceleration due to gravity,

g = - 10

m / s^{2}

(

- ve

as it working downwards)
Let the time taken,

= t

Using 2nd equation of motion,

h = ut + 0.5 g t^{2}

0 = 20 \times t - 5 {\times t}^{2}

5 t = 20

t = 4 s

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