# physic numericals

Time of contact of the ball with floor, Δt = 0.02 sec

Let v1 be the velocity of the ball when dropped before striking the floor,

v2 be the velocity of ball upwards after striking the floor during rebounding

Change in velocity of ball during contact =v2 – (-v1)

so, avg. accln = v2 + v1/ Δt

When a body is dropped, h= 100 m, u=0, a= g, v= v1

from equation, v1^2 = u^2 + 2gh

v1^2 = 2*10*100 = 2000

Consider the motion of ball after striking the floor,

u= v2, v= 0, a= -g, h=20 m, t= 0.02 s

Here the equation of motion becomes

0 = v2^2 + 2*(-10)*20

v2^2 = 400

a = v2 + v1/ 0.02

a = V400 + V2000 / 0.02 (V means underroot)

a = 64.72/ 0.02 = 3236 m/s^2