Abaseball bat contacts a 0.145-kg baseball for 1.3×10−3 s the average force exerted by the bat on the ball is 8900 n. if the ball has an initial velocity of 29 m/s toward the bat and the force of the bat causes the ball's motion to reverse direction, what is the ball's speed as it leaves the bat?

v2 = -50.8m/s

Therefore, the speed of the baseball is 50.8m/s away from the bat. (Reverse direction)

Explanation:

Using the law of conservation of momentum:

The impulse of the bat = change in momentum of the baseball

Ft = ∆M

Ft = m1v1 - m2v2

Since m1 = m2 = m (the mass remains unchanged)

Ft = m(v1 - v2) 1

Given:

Force F = 8900N

Mass m = 0.145kg

time t = 1.3 × 10^-3 s

Initial velocity v1 = 29m/s

Substituting into eqn 1

8900(1.3×10^-3) = 0.145(29-v2)

29-v2 = 79.79

v2 = 29 - 79.79 = -50.8m/s

v2 = -50.8m/s

Therefore, the speed of the baseball is 50.8m/s away from the bat.

that would be because of both gravity and friction. so     c

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation: