A person watching through the window of an apartment sees a ball that rises vertically up then vertically down for a total time of 0.5 sec. If the height of the window is 2m find maximum height above the window reached by the ball?(g=10 m/s^2)

hey santosh,

s=ut+1/2at2

taking the heighest point as the start of the motion, u = 0 and t = 0.5/2 s is the max height reached

= s = 0[(0.5)/2] + 1/2(-10)[(0.5/2)2]

= s = 0 - 2.5/8 = 0.3125

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I will aspect that this will make you understand.

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