Question

A ball is dropped from a tower 40m high. what is its velocity when it has covered 20m? what would be its velocity when it hits the ground?

Answer 1

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HERE IS YOUR ANSWER :-

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S = 20m
S = 1/2 gt^2
20m = 1/2 * 10m/ses * t^2

20 = 5t^2
t^2 = 4
t2 = 2 sec — eq(ii)

s= 40 m
S = ut + 1/2gt^2

S = 1/2*10t^2
40 = 5t^2
t^2 = 8
t1 = 2√2 – eq -(i)

adding eq (I) and (ii)

t1. + t2 = 2√2 sec
t1 + 2 = 2√2
t1 = 2(√2-1)

I) v1 = d/t

= 20/2(√2-1)
= 10/(√2-1) — we rationalised it

= 10(√2+1) m/sec

ii) v = d/t

= 40/2√2
= 10√2 m/sec

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Answer 2

1. When ball fall from 40m and cover 20m…. All the potential energy convert into kinetic energy then

1/2 mv^2 = mgh

V = root of 2gh

Here h = 40 - 20 = 20

V= √(2*10*20)

V= 20m/s

2. When it fall on ground

h=40m

V= √(2*10*40)

V=28.28m/s

You can use g = 9.81 instead of 10. But calcution purpose i use 10 m/s^2