Question

a golf ball is hit from the ground at an angle of 12° with an initial velocity of 35.4 m/s. how long does the ball stay in the air? 0.42 s 1.20 s 2.39 s 7.17 s

Explanation:

Step1: Analyze vertical - motion

The vertical - component of the initial velocity is given by $v_{0y}=v_0\sin\theta$, where $v_0 = 35.4\ m/s$ and $\theta = 12^{\circ}$. So, $v_{0y}=35.4\sin12^{\circ}$.
$v_{0y}=35.4\times0.2079\approx7.36\ m/s$.

Step2: Use vertical - motion equation

The equation for vertical displacement $y - y_0=v_{0y}t-\frac{1}{2}gt^{2}$, when the ball returns to the ground $y - y_0 = 0$. So, $0 = v_{0y}t-\frac{1}{2}gt^{2}=t(v_{0y}-\frac{1}{2}gt)$. One solution is $t = 0$ (corresponds to the initial time). The non - zero solution is obtained by setting $v_{0y}-\frac{1}{2}gt = 0$.
We can solve for $t$: $t=\frac{2v_{0y}}{g}$.
Since $g = 9.8\ m/s^{2}$ and $v_{0y}\approx7.36\ m/s$, then $t=\frac{2\times7.36}{9.8}=\frac{14.72}{9.8}\approx1.50\ s$.

Answer:

$1.50\ s$