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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 :

Answer:

[tex]F = 2.6 \times 10^3 N[/tex]

Explanation:

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

[tex]y = 40 m[/tex]

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

[tex]\frac{1}{2}mv^2 = mgh[/tex]

[tex]v = \sqrt{2gh}[/tex]

[tex]v = \sqrt{2(9.81)(40)}[/tex]

[tex]v = 28 m/s[/tex]

now the change in momentum of the ball is given as

[tex]\Delta P = m(v_f - v_i)[/tex]

[tex]\Delta P = 0.145(28\hat j + 35 \hat i)[/tex]

now force is given as rate of change in momentum

[tex]F = \frac{\Delta P}{\Delta t}[/tex]

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

[tex]F = (1.62 \hat j + 2.03 \hat i)\times 10^3 N[/tex]

so magnitude of the force is given as

[tex]F = \sqrt{1.62^2 + 2.03^2} \times 10^3 N[/tex]

[tex]F = 2.6 \times 10^3 N[/tex]

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