Question

A ball is realesed from a 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

Work done is equal to force*distance.  And, the force is the product of Mass and acceleration.  So, if 'm' is mass, 'g' is the acceleration due to gravity and 'h' is the distance traveled, 'mgh' is the work done.  The distance traveled in each successive second is more than the previous one because the ball continues to accelerate till it reaches its terminal velocity.  Usually, the terminal velocity should be reached in about 10 to 15 seconds of free-fall.

From the formula S = ut + 1/2 at2, calculate h1. h2 and h3 for the first three seconds.  The initial velocity is zero for the first second and for the second and third second, the initial velocity is some positive value.  This value could be calculated using the formula v = u+at.  In both these formulae, a = g.

I have explained the principle involved and leave you to do the calculations.

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Answer 2

Work done is equal to force*distance.  And, the force is the product of Mass and acceleration.  So, if 'm' is mass, 'g' is the acceleration due to gravity and 'h' is the distance traveled, 'mgh' is the work done.  The distance traveled in each successive second is more than the previous one because the ball continues to accelerate till it reaches its terminal velocity.  Usually, the terminal velocity should be reached in about 10 to 15 seconds of free-fall.

From the formula S = ut + 1/2 at2, calculate h1. h2 and h3 for the first three seconds.  The initial velocity is zero for the first second and for the second and third second, the initial velocity is some positive value.  This value could be calculated using the formula v = u+at.  In both these formulae, a = g.

I have explained the principle involved and leave you to do the calculations.