Question

A 0.145-kg baseball pitched horizontally at 35.0 m/s strikes a bat and is popped straight up to a height of 40.0m. If the contact time between bat and ball is 2.5m/s, calculate the magnitude of the average force between the ball and bat during contact. Express your answer using two significant figures.

Answer 1

`F = 2.6 * 10^3 N`

Step-by-step explanation:

Maximum height reached by the ball after being popped by the bat is given as

`y = 40 m`

now we know by energy conservation final speed of the ball after being hit by the bat is given as

`(1)/(2)mv^2 = mgh`

`v = √(2gh)`

`v = √(2(9.81)(40))`

`v = 28 m/s`

now the change in momentum of the ball is given as

`\Delta P = m(v_f - v_i)`

`\Delta P = 0.145(28\hat j + 35 \hat i)`

now force is given as rate of change in momentum

`F = (\Delta P)/(\Delta t)`

`F = (0.145(28\hat j + 35 \hat i))/(2.5 * 10^(-3))`

`F = (1.62 \hat j + 2.03 \hat i)* 10^3 N`

so magnitude of the force is given as

`F = √(1.62^2 + 2.03^2) * 10^3 N`

`F = 2.6 * 10^3 N`