Question

a body thrown vertically up from the ground passes the height of 10.2 minute twice in a couple of 10 second what

Answer 1

hey here your solution

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.
 

Answer 2

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 it will help you mark as brainliest