Mechanics

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 10s. The height h is........... (g=10m/s^2)

Swati

12 Years agoGrade

2 Answers

Arun Kumar

12 Years ago

time of flight 2 vv/g=10

vf=50

=>s=(52^2-50^2)/20

10.2m

Khushi singh

9 Years ago

As given u=52m/s Aplly s=ut+1/2at2s=52×10-1/2×10×10×10s=520-500s=20As body passes the point twice it is the height if bth the points together so to find the height of one pointh=s/2h= 10m

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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 10s. The height h is........... (g=10m/s^2)

Swati

2 Answers

Arun Kumar

Khushi singh

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