# A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat...

## Question:

A pitcher throws a {eq}0.140\ kg {/eq} baseball, and it approaches the bat at a speed of {eq}35.0\ m/s {/eq}. The bat does {eq}W_{nc} = 65.0\ J {/eq} of work on the ball in hitting it. Ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is {eq}25.0\ m {/eq} above the point of impact.

## Work-done:

If the work is done by the object, that means the energy of the object will decrease. And the work-done on the object means that the energy of the object will increase.

Given Data:

• Mass of the baseball: {eq}\rm m \ = \ 0.14 \ kg {/eq}
• Initial speed of the ball: {eq}\rm u \ = \ 35 \ ms^{-1} {/eq}
• Work done by the bat: {eq}\rm W \ = \ 65 \ J \ {/eq}

Kinetic energy of the ball before collision with the bat is

{eq}\rm {K}_i \ = \ \dfrac{1}{2} \times m \times u^2 \ = \ 0.5 \times 0.14 \times 35^2 \ = \ 85.75 \ J {/eq}

Now, after the collision, the total energy of the ball would be

{eq}\rm E_{1} = 85.75 + 65 \\ E_{1} = 150.75 \ J {/eq}

Now, applying the energy conservation

{eq}\begin{align} \rm E_{2} &= \rm E_{1} \\ \rm mgh + 0.5mV^{2} &= \rm E_{1} \\ \rm (0.14 \times 9.8 \times 25) + 0.5(0.14)V^{2} &= 150.75 \\ \rm V &= \rm 40.79 \ m/s \\ \end{align} {/eq}

Where

• h is the height of the ball above the point of impact