Answer to Question #145706 in Physics for Shadie shawnee

Question

A soccer ball is kicked at an angle of 22 degrees and a velocity of 30 m/s. What is the highest point the ball reaches?

Answer 1

Let's write the equation of motion of the soccer ball in vertical direction:

$$y=v_0t_{rise}sin\theta-\dfrac{1}{2}gt_{rise}^2,$$

here, $v_0=30\ \dfrac{m}{s}$ is the initial velocity of the soccer ball, $t_{rise}$ is the time that the soccer ball takes to reach the maximum height (or the highest point), $\theta=22^{\circ}$ is the launch angle, $y$ is the vertical displacement of the soccer ball (or the height) and $g=9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Let's first find the time that the soccer 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 equation for $y$ and find the maximum height:

$$y_{max}=v_0sin\theta\cdot\dfrac{v_0sin\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,$$ $$y_{max}=\dfrac{v_0^2sin^2\theta}{2g},$$ $$y_{max}=\dfrac{(30\ \dfrac{m}{s})^2\cdot sin^222^{\circ}}{2\cdot 9.8\ \dfrac{m}{s^2}}=6.44\ m.$$

Answer:

$y_{max}=6.44\ m.$