Question

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

Answer 1

u need to take the formula u square/2g
which means u= initial velocity and g means gravity then u will get answer 2704/2*10=135.2

Answer 2

$<b><I>As it crosses the point$ twice in its journey then from figure we can see the projectile of motion

Time of accent from point P = Time of decent
then time of decent = 5 sec

now as it will reach its maximum height then
at Hmax , V = 0
then distance between maximum height and point P
-h = 1/2 × -10 × 5
or h = 25 m

now for maximum height
V = 0
then

$from \: {v}^{2} = {u}^{2} - 2gh$
$0 = {52}^{2} - 20s$
or. -2704 = -20 S

or. S = 2704/20

S = 135.2 m

then point P = Hmax - distance between them

= 135.2 - 25

= 112.2 m from ground