Question

question 3 of 10
an athlete jumps with a speed of 4 m/s at an angle of 23°. how long does the athlete stay in the air?
a. 0.75 s
b. 0.32 s
c. 0.92 s
d. 0.55 s

Explanation:

Step1: Find vertical - initial velocity

The initial velocity $v_0 = 4\ m/s$ and the launch angle $\theta=23^{\circ}$. The vertical - initial velocity is given by $v_{0y}=v_0\sin\theta$. So, $v_{0y}=4\sin(23^{\circ})$. Using a calculator, $\sin(23^{\circ})\approx0.391$, then $v_{0y}=4\times0.391 = 1.564\ m/s$.

Step2: Use the kinematic equation for vertical motion

The kinematic equation for vertical displacement $y - y_0=v_{0y}t-\frac{1}{2}gt^2$. When the athlete returns to the same height ($y - y_0 = 0$), the equation becomes $0 = v_{0y}t-\frac{1}{2}gt^2$. Factor out $t$: $t(v_{0y}-\frac{1}{2}gt)=0$. One solution is $t = 0$ (corresponds to the start of the jump). The other non - zero solution is $t=\frac{2v_{0y}}{g}$. Taking $g = 9.8\ m/s^2$, we substitute $v_{0y}=1.564\ m/s$ into the formula: $t=\frac{2\times1.564}{9.8}=\frac{3.128}{9.8}\approx0.32\ s$.

Answer:

B. $0.32\ s$