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?

1.20 s
11.7 s
2.39 s
0.42 s

Explanation:

Step1: Calculate the vertical component of initial velocity

The formula for the vertical component of initial velocity \(v_{0y}\) is \(v_{0y}=v_0\sin\theta\), where \(v_0 = 56.44\ m/s\) and \(\theta = 12^{\circ}\).

$$v_{0y}=56.44\times\sin(12^{\circ})$$

$$v_{0y}=56.44\times0.2079$$

$$v_{0y}\approx11.73\ m/s$$

Step2: Use the kinematic equation for vertical motion

The kinematic equation for vertical displacement \(y = v_{0y}t-\frac{1}{2}gt^{2}\). When the ball hits the ground \(y = 0\). So \(0=v_{0y}t-\frac{1}{2}gt^{2}\), which can be factored as \(t(v_{0y}-\frac{1}{2}gt)=0\). One solution is \(t = 0\) (initial time), and the other is \(t=\frac{2v_{0y}}{g}\) (where \(g = 9.8\ m/s^{2}\)).

$$t=\frac{2\times11.73}{9.8}$$

$$t=\frac{23.46}{9.8}$$

$$t\approx2.39\ s$$

Answer:

2.39 s