physic numericals
a small ball is droped from a height 100m on a flour .the ball rebound to a height of 20m .calculate the average acceleration during the cotact if the ball was in contact of the flour for 0.02 sec take g=10 m/s^2
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
