Question

question 20 of 25
you kick a soccer ball with a speed of 12 m/s at an angle of 21°. how long does it take the ball to reach the top of its trajectory?
a. 2.29 s
b. 1.55 s
c. 0.44 s
d. 0.72 s

Explanation:

Step1: Find initial vertical - velocity

The initial velocity is $v_0 = 12$ m/s and the launch angle is $\theta=21^{\circ}$. The initial vertical - velocity $v_{0y}$ is given by $v_{0y}=v_0\sin\theta$. So, $v_{0y}=12\sin(21^{\circ})$.

Step2: Use kinematic equation for vertical motion

At the top of the trajectory, the vertical velocity $v_y = 0$. The kinematic equation $v_y=v_{0y}-gt$ is used, where $g = 9.8$ m/s². We want to find the time $t$. Rearranging the equation for $t$ gives $t=\frac{v_{0y}-v_y}{g}$. Since $v_y = 0$, then $t=\frac{v_{0y}}{g}=\frac{12\sin(21^{\circ})}{9.8}$.
Calculating $12\sin(21^{\circ})\approx12\times0.3584 = 4.3008$. Then $t=\frac{4.3008}{9.8}\approx0.44$ s.

Answer:

C. 0.44 s