Question

a basketball is launched with an initial speed of 8.0 m/s and follows the trajectory shown. the ball enters the basket 0.96 s after it is launched. what are the distances x and y? note: the drawing is not to scale.

Explanation:

Step1: Find the horizontal - component of the initial velocity

The initial velocity $v_0 = 8.0$ m/s and the launch angle $\theta=45^{\circ}$. The horizontal - component of the initial velocity is $v_{0x}=v_0\cos\theta$. So, $v_{0x}=8.0\cos45^{\circ}=8.0\times\frac{\sqrt{2}}{2}\approx5.66$ m/s.

Step2: Calculate the horizontal distance x

The horizontal motion is a uniform - motion with constant velocity. The formula for horizontal distance is $x = v_{0x}t$. Given $t = 0.96$ s, then $x=5.66\times0.96\approx5.4$ m.

Step3: Find the vertical - component of the initial velocity

The vertical - component of the initial velocity is $v_{0y}=v_0\sin\theta$. So, $v_{0y}=8.0\sin45^{\circ}=8.0\times\frac{\sqrt{2}}{2}\approx5.66$ m/s.

Step4: Calculate the vertical distance y

The formula for vertical displacement in projectile motion is $y=v_{0y}t-\frac{1}{2}gt^{2}$, where $g = 9.8$ m/s². Substitute $v_{0y}\approx5.66$ m/s, $t = 0.96$ s, and $g = 9.8$ m/s² into the formula:
\[

$$\begin{align*} y&=5.66\times0.96-\frac{1}{2}\times9.8\times(0.96)^{2}\\ &=5.4336-4.51584\\ &\approx0.91 \end{align*}$$

\]

Answer:

$x = 5.4$ m, $y = 0.91$ m