Question

A 0.145-kg baseball pitched horizontally at 27 m/s strikes a bat and is popped straight up to a height of 43 m. if the contact time between the bat and the ball is 2.75 ms, calculate the average force [exerted by the bat on the ball] during contact. [let the positive axis lie along the line from the batter to the pitcher, with the batter at the origin.]

Answer 1

The average force exerted by the bat on the ball during contact is -1422.727 N, in the opposite direction of the initial velocity of the ball.

Step-by-step explanation:

To calculate the average force exerted by the bat on the ball during contact, we can use the impulse-momentum principle. The impulse experienced by the ball is equal to the change in momentum. The initial momentum of the ball is the product of its mass and initial velocity, and the final momentum is the product of its mass and final velocity.

Using the given values:

Mass of baseball (m) = 0.145 kg

Initial velocity of baseball (u) = 27 m/s

Final velocity of baseball (v) = 0 m/s (since it's popped straight up)

Contact time (Δt) = 2.75 ms = 0.00275 s

Change in momentum (Δp) = mv - mu

Average force (F) = Δp/Δt

Plugging in the values, we have:

Δp = (0.145 kg)(0 m/s) - (0.145 kg)(27 m/s) = -3.915 kg m/s

F = (-3.915 kg m/s)/(0.00275 s) = -1422.727 N

The negative sign indicates that the force is exerted in the direction opposite to the initial velocity of the ball.

Answer 2

A 0.145-kg baseball pitched horizontally at 27 m/s strikes a bat and is popped straight up to aheight of 43 m. If the contact time between the bat and the ball is