Question

Major league baseball pitchers throw a pitch called a changeup, which looks like a fastball leaving the pitcher's hand but has less speed than a typical fastball. How much more kinetic energy does a 92 mph fastball have than a 78 mph changeup? Express your answer as a percentage of the kinetic energy of the changeup.

Answer 1

percentage of the kinetic energy = 39.1 %

Explanation

V1 = 92 mph

V2 = 78 mph

percentage of kinetic energy = (V1² - V2²) / V2²

percentage of kinetic energy = (92² - 78²) / 78²

percentage of kinetic energy = 0.391

percentage of kinetic energy = 39.1 %

Answer 2

Step 1: Recall the formula for kinetic energy (KE):

$$KE=\frac{1}{2}m{v}^2$$

Since mass ($m$) is the same for both pitches, we can compare their kinetic energies using only their velocities.

Step 2: Write the ratio of the fastball's KE to the changeup's KE:

$$\frac{K{E}_{\text{fastball}}}{K{E}_{\text{changeup}}}=\frac{v_{\text{fastball}}^2}{v_{\text{changeup}}^2}={\left(\frac{92}{78}\right)}^2$$

$${\left(\frac{92}{78}\right)}^2={\left(1.1795\right)}^2\approx 1.39$$

Step 3: Find the percentage difference:

$$\text{Percentage Increase}=(1.39-1)\times 100\mathrm{\%}=39\mathrm{\%}$$

Answer is :

The 92 mph fastball has about 39% more kinetic energy than the 78 mph changeup.