Question

A ball is released from the top of a tower. The ratio of
work done by force of gravity in first second and third
second of the motion of the ball is​

Answer 1

Answer:

Work done =mgh

Since,work done is proportional to distance fall(h).

We know

S=ut+

2

1

gt

2

;here u=0,g=10ms

−2

S=5t

2

if t=1s;S=5.1

2

;S=5

if t=2s;S=5.2

2

−5.1

2

;S=20−5;S=15

if t=3s;S=5.3

2

−5.2

2

;S=45−15;S=25

Hence, the ratio is 1:3:5

Explanation:

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