Question

an acapulco cliff diver in (m = 70.0 kg) jumps 30.0 meters from the top of the cliff toward the ocean below. however, in a freak accident, half - way down 8.00 meters from his launch point, he crashes into a hang glider (total mass 85.0 kg) flying horizontally at 6.5 m/s.
assuming an inelastic collision and terminal velocity from that point on, how far from the intended entry point do the two hit the ocean, and how fast are they moving?
for simplicity, assume no air drag during the divers free fall and a value of g = 10 m/s².

Explanation:

Step1: Calculate the diver's vertical velocity at the collision point

Use the kinematic equation \(v_y^2 = v_{0y}^2+ 2a\Delta y\). The diver starts from rest (\(v_{0y} = 0\)), \(a = g=10\ m/s^2\), and \(\Delta y = 15 - 8=7\ m\).

$$v_y=\sqrt{2g\Delta y}=\sqrt{2\times10\times7}=\sqrt{140}\approx11.83\ m/s$$

Step2: Use conservation of momentum in the horizontal and vertical directions

• Horizontal direction:

Initial momentum \(p_{x,i}=m_{glider}v_{glider}\), final momentum \(p_{x,f}=(m_{diver}+m_{glider})v_x\)

$$v_x=\frac{m_{glider}v_{glider}}{m_{diver}+m_{glider}}=\frac{85\times6.5}{70 + 85}=\frac{552.5}{155}\approx3.56\ m/s$$

• Vertical direction:

Initial momentum \(p_{y,i}=m_{diver}v_y\), final momentum \(p_{y,f}=(m_{diver}+m_{glider})v_y'\)

$$v_y'=\frac{m_{diver}v_y}{m_{diver}+m_{glider}}=\frac{70\times11.83}{155}\approx5.34\ m/s$$

Step3: Calculate the time to fall the remaining vertical distance

The remaining vertical distance \(y = 15\ m\). Use \(y=v_y't+\frac{1}{2}gt^2\) (since \(v_y'\) is the initial vertical velocity after collision and \(a = g\)).

$$15=5.34t + 5t^2$$

Solving the quadratic equation \(5t^2+5.34t - 15=0\) using \(t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\) where \(a = 5\), \(b=5.34\), \(c=- 15\)

$$t=\frac{-5.34\pm\sqrt{(5.34)^2-4\times5\times(-15)}}{2\times5}=\frac{-5.34\pm\sqrt{28.5156 + 300}}{10}=\frac{-5.34\pm\sqrt{328.5156}}{10}$$

$$t=\frac{-5.34\pm18.12}{10}$$

Take the positive root \(t=\frac{- 5.34 + 18.12}{10}=1.28\ s\)

Step4: Calculate the horizontal distance from the intended point

\(x = v_xt\)

$$x=3.56\times1.28\approx4.56\ m$$

Step5: Calculate the final speed

\(v=\sqrt{v_x^2+(v_y'+gt)^2}\)

$$v_y''=v_y'+gt=5.34+10\times1.28 = 18.14\ m/s$$

$$v=\sqrt{3.56^2+18.14^2}=\sqrt{12.67 + 329.06}=\sqrt{341.73}\approx18.49\ m/s$$

Answer:

The horizontal distance from the intended entry point is approximately \(4.56\ m\) and their speed is approximately \(18.49\ m/s\)