Question

a body thrown vertically up with initial velocity 52m/s from the ground passes twice a point at height above at an interval of 10 s. The height h is ? a. 22m b. 10.2m c.11.2m d.15m

Answer 1

A body thrown vertically up with initial velocity 52m/s from the ground passes twice a point at the height above at an interval of 10 s. The height h is?

initial velocity,u = 52 m/s
s=ut+1/2gt²s = 520+1/2(-10)(10)²s = 520-5(10)²s = 520-500=20m

so the height of H is 20m

thankyou

Answer 2

Hello !
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Here 

u = 52 m / s 

t = 10 s 

a = - 10 m / s ²

As we know 

s = ut + 1/2 at²

  = 52 x 10 + (1/2 )(-10) (100)

 = 520  - 500 

= 20 m