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.

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

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 $v\text{'}$.

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

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

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:

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

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

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.

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

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 %.