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

Answer:

$F = 2.6 \times 10^3 N$

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

$\frac{1}{2}mv^2 = mgh$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.81)(40)}$

$v = 28 m/s$

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

$F = \frac{\Delta P}{\Delta t}$

$F = \frac{0.145(28\hat j + 35 \hat i)}{2.5 \times 10^{-3}}$

$F = (1.62 \hat j + 2.03 \hat i)\times 10^3 N$

so magnitude of the force is given as

$F = \sqrt{1.62^2 + 2.03^2} \times 10^3 N$

$F = 2.6 \times 10^3 N$