17. a baseball is hit at 99.4 km/h at 67.6 degrees above the horizontal. if the outfield fence is 120 m…

Explanation:

Step1: Convert the initial velocity to m/s

We know that (1\ km = 1000\ m) and (1\ h=3600\ s). (v_0 = 99.4\ km/h=\frac{99.4\times1000}{3600}\ m/s\approx 27.61\ m/s)

Step2: Find the horizontal and vertical components of the initial velocity

The horizontal component (v_{0x}=v_0\cos\theta), and the vertical component (v_{0y} = v_0\sin\theta), where (\theta = 67.6^{\circ}) (v_{0x}=27.61\times\cos(67.6^{\circ})\approx27.61\times0.381\approx 10.52\ m/s) (v_{0y}=27.61\times\sin(67.6^{\circ})\approx27.61\times0.925\approx25.54\ m/s)

Step3: Find the time of flight

We use the vertical - motion equation (y = v_{0y}t-\frac{1}{2}gt^2). When the ball returns to the initial height ((y = 0)), we have (0=v_{0y}t-\frac{1}{2}gt^2=t\left(v_{0y}-\frac{1}{2}gt\right)). One solution is (t = 0) (initial time). The other solution is (t=\frac{2v_{0y}}{g}) (using (g = 9.8\ m/s^2)) (t=\frac{2\times25.54}{9.8}\approx5.21\ s)

Step4: Find the horizontal range

We use the horizontal - motion equation (x=v_{0x}t) (x=(10.52)\times5.21\approx 54.81\ m)

Answer:

Since the horizontal range (x\approx54.81\ m<120\ m), the hitter does not hit a home - run.