Question

A ball is dropped from the top of a tower.The ratio of work done by force of gravity in 1st ,2nd , and 3rd  second of the motion of ball is​

Answer 1

Answer:

A ball is released from the top of the tower . Then obviously it's acceleration is g . Ball will fall vertically.

Since we know W=mgh

Let us consider W1 for work done by force of gravity in first second

W2 in second second & W3 in third second .

We need to find out h in expression W=mgh

To calculate h , we will use one of the equations of motions i.e

S=ut + 1/2at^2

u=0

H= 1/2gt^2

Now H for first second will be

H= 1/2 g (1)

H for second second will be

H=1/2 g (4)

H for third second will be

H= 1/2g(9)

Now as we have to Know the distance of second second we need to substract the distance of first second.

So required H1 = g/2

H2 = g(4)/2 - g/2 = 3g/2

H3=g(9)/2 - g(4)/2 = 5g/2

Put the values of H in W= mgH

On dividing above 3eqs we get : 1:3:5

So required ratio is 1:3:5

Answer 2

Answer:

The ratio of work done by force of gravity in first, second and third second of motion of the ball is {answer should come 1:3:5} 2) The kinetic energy of a body of mass 2 kg and momentum of 2 Ns is: {answer should come 1 joule}

Explanation:

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