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A body dropped from certain height h reaches the ground after certain time t .after half of the time its height above ground is given 3h/4 bt hw ?explain?



The distance traveled by an accelerating body is given by S=U*t+ (0.5*a*t^2). where U is the initial velocity, a is the acceleration of the body and t is the time. So in case of free falling body, U=0, a=9.8 m/s2. It T is the time taken to reach the ground, i.e to travel the distance h, then the distance traveled in T/2 is 
H=(0.5*a* T^2)/4= h/4. This implies the distance of the object above the ground is h-h/4=3h/4
0 0 0
H=1/2 X a X t²/4=h/4
height from ground=h-h/4=3h/4
1 5 1
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