Question

a golf ball leaves a tee at 60m/s and strikes the ground 200 m away. At what two angles with the horizontal could it have begun its flight? Find the time of flight and maximum altitude in each case.

Answer 1

Question 15512

From one side, one knows that

$$v _ {x} = v _ {0} \cos \varphi , S _ {x} = T v _ {0} \cos \varphi = 2 0 0 m$$

For vertical motion,

$$S _ {y} = v _ {0} t \sin \varphi - \frac {g t ^ {2}}{2}$$

The ball reaches the maximum altitude, when $\frac{dS_y}{dt} = 0, \Rightarrow t_{1/2} = \frac{v_0\sin\varphi}{g}$ - this gives the half of the time of the flight, so the full time of the flight is

$$T = \frac {2 v _ {0} \sin \varphi}{g}$$

Now using (1) and (3), obtain: $\sin 2\varphi = \frac{S\cdot g}{v_0^2}$ , which gives $\varphi_{1}\approx 16.51$ for $\varphi \in [0,\pi /2]$ . The second angle, is obviously $\varphi_{2} = \pi -2\varphi_{1}\approx 146.98$ (this will give the motion to the left, if for $\varphi_{1}\approx 16.51$ the motion was to the right). The time of the flight and maximum altitude will be equal for both angles. Hence,

using (3), $T = \frac{2v_0\sin\varphi}{g}\approx 3.48s$ , and

using (2), $h_{\text{max}} = S_y|_{t = t_{1/2}} = \left[ v_0 t \sin \varphi - \frac{g t^2}{2} \right] |_{t = t_{0.5}} = \frac{v_0^2 \sin^2 \varphi}{2 g} \approx 14.82 \, m$ .