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)
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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
which means u= initial velocity and g means gravity then u will get answer 2704/2*10=135.2
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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
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
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
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
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