# 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)

u = 52 m/s
s = ut + $\frac{1}{2}$gt2
s = 520 + $\frac{1}{2}$($-$10)(10)2
s = 520 $-$ (5)(10)2
s = 520 $-$ 500 = 20 m
Given that
s = 2H
2H =20 m
$⇒$H = 10 m

• 35

s=ut-1/2*gt2

s=52*10- 5(10)2

s=520-500

s=20m

As it passes twice a point at H height.

i.e 2H = 20m

H=10m

IF I AM RIGHT THUMBS UP BUDDY

• 28
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