a ball released from certain height the ground in time T. where it will be from the point of release at time 5T/6(from the top)
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in the given case, we have, ... and u is 0 as in the beginning the ball was "dropped" ... Hence the ball falls a distance of h/4 in time T/2, i.e. the ball will be at a ...
vivek583:
please answer fully
The ball's displacement when it reaches the ground in time T is h.
Hence,
s = h, t = T, a = g, u = 0
Here s is displacement of the body, in this case ball, in time t moving with acceleration a which in this case is acceleration due to gravity g, u is initial velocity given to the body at the start of the displacement, which is 0 in this case because the ball is dropped and not thrown.
Note: All the measurements are made wrt a frame of reference in
Hence,
s = h, t = T, a = g, u = 0
Here s is displacement of the body, in this case ball, in time t moving with acceleration a which in this case is acceleration due to gravity g, u is initial velocity given to the body at the start of the displacement, which is 0 in this case because the ball is dropped and not thrown.
Note: All the measurements are made wrt a frame of reference in
in which downward direction is measured positive.
According to equation of motion
s = ut + (1/2)at²
h = 0 + (1/2)gT²
T² = 2h/g
Let us apply the same equation of motion for displacement after time T/2.
s = ut + (1/2)at²
s = 0 + (1/2)g(T/2)²
s = (1/2)(g/4)T²
Substituting the value of T²
s = (1/2)(g/4)(2h/g)
s = h/4
Hence the ball falls a distance of h/4 in time T/2, i.e. the ball will be at a height of 3h/4 or 0.75h after time T/2.
According to equation of motion
s = ut + (1/2)at²
h = 0 + (1/2)gT²
T² = 2h/g
Let us apply the same equation of motion for displacement after time T/2.
s = ut + (1/2)at²
s = 0 + (1/2)g(T/2)²
s = (1/2)(g/4)T²
Substituting the value of T²
s = (1/2)(g/4)(2h/g)
s = h/4
Hence the ball falls a distance of h/4 in time T/2, i.e. the ball will be at a height of 3h/4 or 0.75h after time T/2.
sorry please tell clearly i am 8class
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