Question

1. a golf ball is hit from the ground at an angle of 12° with an initial velocity of 56.44 m/s. how long does the ball stay in the air? 0.42 s 1.20 s 2.39 s 11.7 s

Explanation:

Step1: Identify vertical - initial velocity

The initial velocity $v_0 = 56.44$ m/s and the launch angle $\theta=12^{\circ}$. The vertical - component of the initial velocity is given by $v_{0y}=v_0\sin\theta$. So, $v_{0y}=56.44\times\sin(12^{\circ})$.
$v_{0y}=56.44\times0.2079 = 11.73$ 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 ball returns to the ground $y - y_0 = 0$. So, $0 = v_{0y}t-\frac{1}{2}gt^2$. Factor out $t$: $t(v_{0y}-\frac{1}{2}gt)=0$. One solution is $t = 0$ (corresponds to the initial time when the ball is hit). The other non - zero solution is obtained by setting $v_{0y}-\frac{1}{2}gt = 0$. Solving for $t$ gives $t=\frac{2v_{0y}}{g}$.
We know that $g = 9.8$ m/s² and $v_{0y}=11.73$ m/s. Then $t=\frac{2\times11.73}{9.8}=\frac{23.46}{9.8}=2.39$ s.

Answer:

$2.39$ s