Question

A body thrown vertically up from the ground passes the height of 10.2m twice in an interval of 10 seconds what was the initial velocity

Answer 1

Answer:

Here, g=10 m/s^2 , v = 0 m/s (velocity at Hmax

Since, v=u- gt

=>   0 = u-gt

=>   u = gt

therefore, speed of the body at 10.2 m height is then 

u = 10 * 5

u  = 50 m/s

Let distance between Hmax and height at 10.2 m be x 

using s = ut -0.5at^2

     

=>   x = 50*5 - 0.5*10*5*5

=>  x = 125m

therefore, Hmax = 125+10.2 m

                        = 135.2 m

but, Hmax = u^2/2g

=> 135.2 = u^2/2*10

=> u = 52 m/s

therefore, Initial velocity of the body is 52 m/s