Question

A basketball is shot from an initial height of 2.40 m (Fig. 3-57) with an initial speed$v_0=12\mathrm{m}/\mathrm{s}$directed at an angle${\theta }_0={35}^o$above the horizontal. (a) How far from the basket was the player if he made a basket? (b) At what angle to the horizontal did the ball enter the basket?

FIGURE 3-57Problem 71

Short answer

(a) The distance between the player and the basket is 12.78 m.

(b) The ball enters the basket at an angle of $-30.79°$from the horizontal. (Negative sign indicates that angle is measured in a clockwise direction.)

Step-by-step solution

Step 1. Angle of landing

The angle of landing with the horizontal axis will be equal to the angle the velocity vector makes with the horizontal axis.

Step 2. Given data and assumptions

The initial velocity with which the basketball is launched,$v_0=12\mathrm{m}/\mathrm{s}$

The initial vertical distance from the ground,$y_{\mathrm{i}}=2.40\mathrm{m}$

The final vertical distance from the ground,$y_{\mathrm{f}}=3.05\mathrm{m}$

The angle of projection of the basketball,$\theta =35°$

Let the distance of the player from the basket be x, and the time taken by the basketball to reach the basket be t.

Calculating the distance of the player from the basket

Setting up equations for vertical motion

The initial velocity along the vertical direction will be:

$v_{\mathrm{y}}=v_0\sin \theta$.

Applying the second equation of motion for the vertical direction, you get:

$y_{\mathrm{f}}-y_{\mathrm{i}}=v_{\mathrm{y}}t+\frac{1}{2}\left(g\right)t^2$

Substituting the values in the above equation, you get:

$\begin{array}{c}3.05-2.40=12\sin 35°t+\frac{1}{2}\left(-9.8\right)t^2\\ 4.9t^2-6.88t+0.65=0\end{array}$

The roots of the above quadratic equation are t = 1.30 s and t = 0.10 s. The lower time represents the time the ball reaches the vertical height $y_{\mathrm{f}}$when it is rising upward, and the higher time t = 1.30 s represents the time the ball reaches the basket.

Thus, the time at which the ball reaches the basket is t = 1.30 s.

Setting up equations for horizontal motion

The initial velocity along the horizontal direction will be:

$v_{\mathrm{x}}=v_0\cos \theta$.

Applying the second equation of motion for the vertical direction, you get:

$x=v_0\cos \theta t+0$

(As acceleration in the horizontal direction is zero)

Substituting the values in the above equation, you get:

$x=12\cos 35°\times 1.30$

Solving for x from the above equation, you get:

x = 12.78 m

Thus, the horizontal distance between the player and the basket is 12.78 m.

Step 4. Calculating the angle at which the ball hits the basket

Let the angle at which the ball hits the basket be $\theta$. We know that the acceleration along the horizontal direction of motion is zero; thus, the velocity along the horizontal direction will not change and will be equal to $v_{\mathrm{x}}$.

The velocity when the ball is about to hit the basket can be calculated by using the first equation of motion as:

$v_{\mathrm{yf}}=v_{\mathrm{y}}+ \mathrm{g}t$

($v_{\mathrm{yf}}$ is the velocity in the vertical direction just before hitting the basket.)

Substituting the values in the above equation, you get:

$\begin{array}{c}{v}_{\mathrm{yf}}=12\sin 35°-9.8\times 1.30\\ =-5.86\mathrm{m}/\mathrm{s}\end{array}$

The negative sign implies that the velocity is in the negative direction.

The tangent function of the angle at which the ball hits the basket is:

$\begin{array}{c}\tan \theta =\frac{v_{\mathrm{yf}}}{v_{\mathrm{x}}}\\ =\frac{-5.86}{12\cos 35°}\end{array}$

Solving the above equation for\[\theta \], you get:

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-5.86}{12\cos 35°}\right)\\ =-30.79°\end{array}$

Thus, the trajectory of the ball makes an angle of $-30.79°$with the horizontal axis passing through the basket. (The negative sign indicates that the angle is measured in a clockwise direction below the axis.)