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