Question

a+bal+is+dropped+from+an+height+10m+and+mass+of+ball+is+0.5kg+how+many+time+it+will+take+to+come+at+rest+and+how+much+total+distance+travelled by a ball

Answer 1

Initial height (h) = 10m

initial velocity (u) = 0

velocity just before hotting ground = √2gh = √2*9.8*10 = 14.07 m/s (downward)

after colliding,

maximum height reached = 2.5m

final velocity (v) at top = 0

initial velocity(just after collision) = √2gh = √2*9.8*2.5 = √49 = 7 m/s (upward)

assuming downward as positive direction,

So velocity just before hitting ground = +14.07 m/s 

velocity just after hitting ground = -7 m/s 

change in velocity = +14.07 - (-7) = 21.07 m/s

time = 0.01s

acceleration = change in velocity/time = 21.07/0.01 = 2107 m/s²

approximately = 2100 m/s²

P.S. Sorry if there is any sort of mistake.