A ball is thrown upwards .the distance covered by it during of last 2 second before it reaches the maximum height is ???? Find answer
v at top = 0
v at two seconds down = Vi
0 = Vi - g t
so
Vi = 9.81 * 2
average speed during those two seconds is
Vaverage = (0 + 2*9.81) /2
so average speed = g = 9.81
9.81 m/s * 2 s = 19.6 meters
By the way that was quite a throw :)