Question

23 if a weight is launched with an initial velocity of 45.9 meters per second, at an angle of 66.1° above horizontal, how much time will it take the balloon to reach the ground again? 8.10 s 8.56 s 4.14 s 3.79 s

Explanation:

Step1: Find 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 \(v_0 = 45.9\space m/s\) and \(\theta = 66.1^{\circ}\).

$$v_{0y}=45.9\times\sin(66.1^{\circ})$$

$$v_{0y}=45.9\times0.914$$

$$v_{0y}\approx41.9\space m/s$$

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 weight reaches the ground again, \(y - y_0 = 0\). So, \(0 = v_{0y}t-\frac{1}{2}gt^2\). Factoring out \(t\) gives \(t(v_{0y}-\frac{1}{2}gt)=0\). One solution is \(t = 0\) (initial time). The other solution is \(t=\frac{2v_{0y}}{g}\), where \(g = 9.8\space m/s^2\).

$$t=\frac{2\times41.9}{9.8}$$

$$t=\frac{83.8}{9.8}$$

$$t\approx8.55s\approx8.56s$$

Answer:

8.56 s