A body thrown vertically up with initial velocity 52m/sec.from the ground passes twice a point at h height above at an interwal of 10sec the height h is?
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Given,
the body was thrown up with an initial velocity, u = 52 m/sit passes twice a point of height h in t= 10 sec.
s= ut + 1/2 at²
= 52*10 - 1/2*10 * (10)²
= 520-500
= 20m
it passes twice a point of Height,
S(distance) =2 h = 20 m ( h = height)
2 h = 20 m
h = 10 m
The height h= 10 m
the body was thrown up with an initial velocity, u = 52 m/sit passes twice a point of height h in t= 10 sec.
s= ut + 1/2 at²
= 52*10 - 1/2*10 * (10)²
= 520-500
= 20m
it passes twice a point of Height,
S(distance) =2 h = 20 m ( h = height)
2 h = 20 m
h = 10 m
The height h= 10 m
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Explanation:ans 10.2
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