A ball is drop from a top of tower of height 'h'. what time does it take to reach the ground? what will be it speed at the time of striking?
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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 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 0.75hafter time T/2.
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 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 0.75hafter time T/2.
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