A tennis ballis dropped on to the floor from a height of 4m.It rebounds to a height of 2m.If the ball is in contact with the floor for 12xx10^(-3)s, its average acceleration during

A tennis ballis dropped on to the floor from a height of 4m.It rebound

1.	A tennis ballis dropped on to the floor from a height of 4m.It rebounds to a height of 2m.If the ball is in contact with the floor for 12xx10^(-3)s, its average acceleration during
Answer» 0 `1260ms^(-2)` `980ms^(-2)` `600ms^(-2)` Solution : `V^(2)=u^(2)+2as` As u=0 , HENCE `V^(2)= 2as` or `V = sqrt2as` When the tennis ball is DROPPED the velocity gained by it just before colliding with the floor is : `u= sqrt2xx9.8xx4 = 8.85ms^(-1)` (upwords) `=- 8.85ms^(-1)` (upwords) time of contact of the ball with the floor (GIVEN `t = 12xx10^(-3)S` Hence , average acceleration `a= (v-u)/(t) = (6.26-(-8.85)/(12xx10^(-3)` `=+1260ms^(-2)`