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A ball after freely falling from a height of 4.9 m, hits a horizontal surface. If e=3/4, Then the ball will hit the surface, second time after ;a} 1sec b} 1.5secc}2secd}3sec

ravi , 3 Years ago
Grade 11
anser 1 Answers
Ayush Kumar

Last Activity: 3 Years ago

Hello dear
The question which you are asking is a bit related to kinematics and a bit related to collision
Just remeber some equations i.e.
Time period of an object=2u/g where u is the initial velocity
Velocity of an object when it reaches the ground9if it is going under free fall)=root under 2gh
e=velocity of seperation/velocity of approach
Now,
Velocity of approach =root under 2*9.8*4.9=9.8
e=3/4
Velocity of seperation=3/4*9.8
Time period=2*u/g
now u=velocity of seperation=3/4*9.8
T=2*3*9.8/4*9.8
T=1.5sec

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