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?

Explanation:

Step1: Analyze vertical motion

The vertical displacement \(y = - 87\ m\) (taking downwards as negative), the initial vertical velocity \(v_{0y}=v_{0}\sin\theta\) where \(v_{0} = 44.8\ m/s\) and \(\theta = 12^{\circ}\), and the acceleration \(a=-g=- 9.8\ m/s^{2}\). Use the equation \(y = v_{0y}t+\frac{1}{2}at^{2}\).
Substitute \(v_{0y}=44.8\sin12^{\circ}\approx44.8\times0.208 = 9.32\ m/s\) into \(y = v_{0y}t+\frac{1}{2}at^{2}\), we get \(-87=9.32t-4.9t^{2}\).

Step2: Solve the quadratic equation

The quadratic equation \(4.9t^{2}-9.32t - 87=0\). Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 4.9\), \(b=-9.32\), \(c = - 87\).
First calculate the discriminant \(\Delta=b^{2}-4ac=(-9.32)^{2}-4\times4.9\times(-87)=86.86 + 1695.6=1782.46\).
Then \(t=\frac{9.32\pm\sqrt{1782.46}}{9.8}\). We take the positive root \(t=\frac{9.32+\sqrt{1782.46}}{9.8}\approx\frac{9.32 + 42.22}{9.8}=\frac{51.54}{9.8}\approx5.26\ s\).

Step3: Calculate horizontal distance

The initial horizontal velocity \(v_{0x}=v_{0}\cos\theta\), \(v_{0}=44.8\ m/s\), \(\theta = 12^{\circ}\), so \(v_{0x}=44.8\cos12^{\circ}\approx44.8\times0.978 = 43.8\ m/s\).
Using the formula \(x = v_{0x}t\), substitute \(v_{0x}\approx43.8\ m/s\) and \(t\approx5.26\ s\), we get \(x\approx43.8\times5.26\approx230\ m\).

Answer:

The ball travels approximately \(230\ m\) horizontally before plunging into the ocean.