Question

If a basketball pitch leaves the pitcher’s hand horizontally at a velocity of$\mathbf{150}\mathbf{km}/\mathbf{h}$, by what % will the pull of gravity change the magnitude of the velocity when the ball reaches the batter, 18 m away? For this estimate, ignore air resistance and spin on the ball.

Short answer

When the ball would reach the batter, the magnitude of its velocity would have been increased by 2.90 %.

Step-by-step solution

Step 1. Calculation of magnitude of a vector using its rectangular components

The magnitude of a vector is equal to the square root of the sum of the square of its rectangular components.

Step 2. Given data and assumptions

The initial velocity with which the ball is launched,$v_0=150\mathrm{km}/\mathrm{h}$

The distance between the pitcher and the batter,$d=18\mathrm{m}$

Let the magnitude of the final velocity of the ball when it reaches the batter be $v\text{'}$.

Step 3. Converting the unit of velocity

The velocity in $\mathrm{m}/\mathrm{s}$can be written as:

$\begin{array}{c}{v}_0=\left(150\mathrm{km}/\mathrm{h}\right)\times \frac{1000\mathrm{m}/\mathrm{km}}{3600\mathrm{s}/\mathrm{h}}\\ =41.67\mathrm{m}/\mathrm{s}\end{array}$

Step 4. Calculating the components of the initial velocity

As the ball is launched horizontally, the component in the vertical direction will be zero.

The component in the horizontal direction is:

$v_{\mathrm{x}}=v_0$

The component in the vertical direction is:

$v_{\mathrm{y}}=0$

Step 5. Calculating the time taken in traveling the horizontal distance of 18 m

Applying the second equation of motion for horizontal motion, you get:

$d=v_{\mathrm{x}}t+0$

(As acceleration in the horizontal direction is zero)

Substituting the values in the above equation, you get:

$18=41.67\times t$

Calculate the value of t.

t = 0.43 s

Step 6. Calculating the vertical and horizontal components of velocity when the ball reaches the batter

The vertical component of velocity at the time when the ball reaches the batter can be calculated by the first equation of motion as:

$\begin{array}{c}{v\text{'}}_{\mathrm{y}}=0+gt\\ =-9.8\times 0.43\\ =-4.214\mathrm{m}/\mathrm{s}\end{array}$

As there is no acceleration in the horizontal direction, the velocity in the horizontal direction will not change.

$\begin{array}{c}{v\text{'}}_{\mathrm{x}}=v_0\\ =41.67\mathrm{m}/\mathrm{s}\end{array}$

Step 7. Calculating the percentage change in the magnitude of the velocity

The magnitude of velocity at the end can be written in the form of its horizontal and vertical components as:

$v\text{'}=\sqrt{{\left({v\text{'}}_{\mathrm{x}}\right)}^2+{\left({v\text{'}}_{\mathrm{y}}\right)}^2}$

Substitute the values in the above equation.

$\begin{array}{c}v\text{'}=\sqrt{{4.214}^2+{41.67}^2}\\ =41.88\mathrm{m}/\mathrm{s}\end{array}$

The percentage change in the magnitude will be calculated as:

role="math" localid="1644997757938" $\begin{array}{c}\%\mathrm{change}=\frac{v\text{'}-v_0}{v_0}\times 100\\ \frac{42.88-41.67}{41.67}\times 100\\ =2.90\%\end{array}$

Thus, the percentage change in magnitude will increase by 2.90 %.