Question

A bat moving at 90 km/h strikes an oncoming ball moving at 110 km/h. If the batter follows through so that the speed of the bat doesn't change during the hit and the coefficient of restitution of the ball is 0.6, what speed does the batted ball have after it has been struck?

Answer 1

Answer:

The final speed of the ball will be 210 km/h.

Explanation:

Given that

Speed of bat = 90 km/h

$u_1=90\ km/h$

Speed of ball= 110 km/h

$u_2=110\ km/h$

Speed of the bat will remain constant

$v_1=90\ km/h$

Lets speed of the ball after striking the bat is $v_2$

As we know that coefficient of restitution given as

$e=\dfrac{v_2-v_1}{u_1-u_2}$

$0.6=\dfrac{v_2-90}{90+100}$

Both bat and ball are moving in opposite direction before the collision.

$0.6=\dfrac{v_2-90}{90+100}$

So we can say that

$v_2=210\ km/h$

The final speed of the ball will be 210 km/h.