a body thrown vertically up from the ground passes the height 10.2 m twice at an interval of10s. what was its intial velocity​

Hi Rohini,

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.


Hope this is helpful.