Question

A golfer is tryinh to improve the range of her shot. To do so she drives a golf ball from the top of a steer cliff 30.0m above the ground where the ball will land. If the ball has an initial velocity of 25m/s and is launched at an angle at an angle of 50 degree above the horizontal, determine the ball's time of flight,its range, and its final just before it hits the ground

Answer 1

Horizontal speed: $v_h = v \cdot \cos \alpha$. Vertical speed: $v_v = v \cdot \sin \alpha$.

So, we will have:

$$\begin{array}{l} v_v - g \cdot t_1 = 0 \Rightarrow t_1 = \frac{v_v}{g} \Rightarrow \\ \Rightarrow t_1 = \frac{25 \cdot \sin 50{}^\circ}{9.8} \approx 1.95 \text{ s}. \end{array}$$ $$s = 30 \text{ m}$$ $$s = v_v \cdot t_2 + \frac{g \cdot t_2^2}{2}$$ $$\begin{array}{l} 30 = 25 \cdot \sin 50{}^\circ \cdot t_2 + \frac{9.8 \cdot t_2^2}{2} \Rightarrow 4.9 \cdot t_2^2 + 19.15 \cdot t_2 - 30 = 0 \Rightarrow \\ \Rightarrow t_2 = \frac{-19.15 \pm \sqrt{954.765}}{9.8} \Rightarrow t_2 = 5.1 \text{ s}. \end{array}$$ $$t = 2 \cdot t_1 + t_2 = 2 \cdot 1.95 + 5.1 = 9 \text{ s}.$$

So, we have the distance:

$$d = v_h \cdot t = 9 \cdot 25 \cdot \cos 50{}^\circ \approx 145 \text{ m}.$$

Answer: 145 meters.