Question

problem 7: a basketball players shoots a ball. the initial speed of the ball is 6.9 m/s at an angle of 54° from the positive horizontal direction. the ball is released 2 m high from the ground. after a time of = 1.2 s, find the final position of the ball, where r0 = (0,2) m, assuming only gravity is acting on the ball after release. do you think the player scored? explain.

Explanation:

Step1: Find initial horizontal and vertical velocities

The initial horizontal velocity $v_{0x}=v_0\cos\theta$, and initial vertical velocity $v_{0y}=v_0\sin\theta$. Given $v_0 = 6.9\ m/s$ and $\theta = 54^{\circ}$.
$v_{0x}=6.9\cos(54^{\circ})\approx 6.9\times0.5878 = 4.0558\ m/s$
$v_{0y}=6.9\sin(54^{\circ})\approx 6.9\times0.8090 = 5.6821\ m/s$

Step2: Calculate horizontal position

The horizontal - motion is a uniform - motion with $x = x_0+v_{0x}t$. Since $x_0 = 0$ and $t = 1.2\ s$, $x=0 + 4.0558\times1.2=4.86696\ m$

Step3: Calculate vertical position

The vertical - motion is a uniformly - accelerated motion with $y=y_0+v_{0y}t-\frac{1}{2}gt^2$, where $y_0 = 2\ m$, $v_{0y}=5.6821\ m/s$, $t = 1.2\ s$ and $g = 9.8\ m/s^2$.
$y=2+5.6821\times1.2-\frac{1}{2}\times9.8\times(1.2)^2$
$y=2 + 6.81852-7.056$
$y=1.76252\ m$

The standard height of a basketball hoop is $3.05\ m$. Since the height of the ball at $t = 1.2\ s$ is $y = 1.76252\ m$ which is less than the height of the hoop, and we don't know the horizontal distance to the hoop precisely (but assuming a normal basketball - shooting scenario, the horizontal distance is usually not enough to reach the hoop at this height), the player likely did not score.

Answer:

The final position of the ball is $(4.87\ m,1.76\ m)$ (rounded to two decimal places). The player likely did not score.