Question

A basketball player wishes to make a stunning last-second basket. With .8 seconds on the clock, she wants the ball to enter the hoop (height 3.05m) as time expires in the game. She shoots the ball with an initial speed of 7 m/s, from 2.1 meters above the ground.
1. At what angle (from the HORIZONTAL) does she shoot the ball?

2. Draw qualitatively accurate position, velocity, and acceleration graphs for both dimensions of the ball's motion while it is in flight.

Answer 1

Answer:

(1) $46.86^\circ$

(2) Diagram has been attached in the solution.

Explanation:

This question is from projectile motion.

From the given question, we will discuss the motion of the basket ball only in the vertical direction from which we will be able to find out the angle of the initial velocity with the horizontal with which it should be shoot to enter the hoop.

Part (1):

Let us assume:

• $y_i$ = initial position of the basket ball = 2.1 m
• $y_f$ = final position of the basket ball = 3.05 m
• $a_y$ = acceleration of the ball along the vertical = $-9.8\ m/s^2$
• $t$ = time taken to reach the goal = 0.8 s
• $\theta$ = angle of the initial velocity with the horizontal
• $u$ = initial speed of the ball = 7 m/s
• $u_y$ = initial vertical velocity of the ball = u\sin \theta

Using the equation of motion for constant acceleration, we have

$y_f-y_i=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 3.05-2.1=u\sin \theta (0.8) +\dfrac{1}{2}\times (-9.8)(0.8)^2\\\Rightarrow 0.95=7\times \sin \theta (0.8) -3.136\\\Rightarrow 0.95=5.6\sin \theta -3.136\\\Rightarrow 5.6\sin \theta= 0.95+3.136\\\Rightarrow 5.6\sin \theta= 4.086\\\Rightarrow \sin \theta= \dfrac{4.086}{5.6}\\\Rightarrow \sin \theta=0.729\\\Rightarrow \theta=\sin^{-1}(0.729)\\\Rightarrow \theta=46.86^\circ$

Hence, the angle of the shoot of the basket ball with the horizontal is $46.86^\circ$ such that it reaches the hoop on time.

Part (2):

For this part, a diagram has been attached.