Question

a man stands at the edge of a cliff and hits a golf ball into the ocean 87 m below. he hits the ball 44.8 m/s at an angle of 12 degrees above horizontal. ignoring air resistance, how far does the ball travel horizontally before plunging into the ocean?

231 m

41.6 m

116 m

83.3 m

Explanation:

Step1: Calculate the initial vertical and horizontal velocities

The initial velocity \(v_0 = 44.8\space m/s\) and the launch angle \(\theta=12^{\circ}\).
The initial horizontal velocity \(v_{0x}=v_0\cos\theta\), so \(v_{0x}=44.8\cos(12^{\circ})\approx44.8\times0.9781 = 43.8\space m/s\).
The initial vertical velocity \(v_{0y}=v_0\sin\theta\), so \(v_{0y}=44.8\sin(12^{\circ})\approx44.8\times0.2079 = 9.31\space m/s\).

Step2: Use the vertical displacement formula to find the time of flight

The vertical displacement \(y - y_0=- 87\space m\) (taking downwards as negative), and the vertical motion equation is \(y - y_0=v_{0y}t-\frac{1}{2}gt^2\), where \(g = 9.8\space m/s^2\).
Substituting the values: \(-87=9.31t-4.9t^2\).
Rearranging to the quadratic form \(4.9t^2-9.31t - 87=0\).
Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\), where \(a = 4.9\), \(b=-9.31\), \(c=-87\).
\(t=\frac{9.31\pm\sqrt{(-9.31)^2-4\times4.9\times(-87)}}{2\times4.9}=\frac{9.31\pm\sqrt{86.68 + 1699.2}}{9.8}=\frac{9.31\pm\sqrt{1785.88}}{9.8}=\frac{9.31\pm42.26}{9.8}\).
We take the positive root \(t=\frac{9.31 + 42.26}{9.8}=\frac{51.57}{9.8}\approx5.26\space s\).

Step3: Calculate the horizontal distance

Using the formula \(x=v_{0x}t\).
Substituting \(v_{0x}\approx43.8\space m/s\) and \(t = 5.26\space s\), we get \(x=43.8\times5.26\approx231\space m\).

Answer:

231 m