a body thrown vertically up with initial velocity 52 m/sec from the ground passes twice a point at H height - Physics - Motion In A Plane

Question

a body thrown vertically up with initial velocity 52 m/sec from
the ground passes twice a point at H height above at an interval of
10 s The height H is (Take g= 10)

Ankit Pasricha · Accepted Answer

u = 52 m/s
s = ut + 12gt2
s = 520 + 12(-10)(10)2
s = 520 - (5)(10)2
s = 520 - 500 = 20 m
Given that
s = 2H
2H =20 m
⇒H = 10 m

a body thrown vertically up with initial velocity 52 m/sec from the ground passes twice a point at H height above at an interval of 10 s . The height H is (Take g= 10)