Answer to Question #142079 in Physics for Inday

Question

1. A STEM student is trying to estimate the average velocity of the players in a basketball team by studying the previous game of the players. One particular player can produce an initial velocity of 5 m/s . If the player throws a basketball from a height of 2 meters at an angle of 45 degrees with respect to the horizontal axis and the given average velocity of 5 m/s , calculate: a. The maximum height reached by the ball; b. The maximum horizontal range to reach the same level as it was thrown; and c. The time for the ball to reach its maximum height

Answer 1

a) Let's write the equations of motion of the ball in horizontal and vertical directions:

$$x=v_0tcos\theta, (1)$$ $$y=y_0+v_0tsin\theta-\dfrac{1}{2}gt^2, (2)$$

here, $x$ is the horizontal displacement of the ball (or the range of the ball), $v_0=5.0\ \dfrac{m}{s}$ is the initial velocity of the ball, $t$ is the total flight time of the ball, $\theta=45^{\circ}$ is the launch angle, $y$ is the vertical displacement of the ball (or the height), $y_0=2.0\ m$ is the initial height from which the ball was thrown and $g=9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Let's first find the time that the ball takes to reach the maximum height from the kinematic equation:

$$v_y=v_0sin\theta-gt_{rise},$$ $$0=v_0sin\theta-gt_{rise},$$ $$t_{rise}=\dfrac{v_0sin\theta}{g}.$$

Then, we can substitute $t_{rise}$ into the second equation and find the maximum height:

$$y_{max}=y_0+v_0sin\theta\cdot\dfrac{v_0sin\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,$$ $$y_{max}=y_0+\dfrac{v_0^2sin^2\theta}{2g},$$ $$y_{max}=2.0\ m+\dfrac{(5.0\ \dfrac{m}{s})^2\cdot sin^245^{\circ}}{2\cdot 9.8\ \dfrac{m}{s^2}}=2.64\ m.$$

b) Let's first find the time the ball takes to reach the same level as it was thrown:

$$t=2t_{rise}=\dfrac{2v_0sin\theta}{g}.$$

Then, we can substitute $t$ into the first equation and find the maximum horizontal range to reach the same level as the ball was thrown:

$$x=v_0cos\theta\cdot \dfrac{2v_0sin\theta}{g}=\dfrac{v_0^2sin2\theta}{g},$$ $$x=\dfrac{(5.0\ \dfrac{m}{s})^2\cdot sin2\cdot45^{\circ}}{9.8\ \dfrac{m}{s^2}}=2.55\ m.$$

c) Finally, we can find the time that the ball takes to reach the maximum height from the formula:

$$t_{rise}=\dfrac{v_0sin\theta}{g}=\dfrac{5.0\ \dfrac{m}{s}\cdot sin45^{\circ}}{9.8\ \dfrac{m}{s^2}}=0.36\ s.$$

Answer:

a) $y_{max}=2.64\ m.$

b) $x=2.55\ m.$

c) $t_{rise}=0.36\ s.$