Answer to Question #139782 in Physics for bbjj

Question

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

Answer 1

Equations of motion of a ball

$$x(t)=v_0\cos\theta t =16.1t$$ $$y(t)=y_0+v_0\sin\theta t -\frac{gt^2}{2}=30.0+19.2t-4.9t^2$$

(a) when a ball reaches the ground, then $y(t)=0,$ so

$$30.0+19.2t-4.9t^2=0$$

Root

$$t_{\star}=5.12\: s$$

(b) The range

$$R=x(t_{\star})=x(5.12)=16.1\times 5.12=82.4\: \rm m$$

(c) The final velocity

$$v=\sqrt{v_0^2+2gy_0}\\ =\sqrt{25.0^2+2\times 9.8\times 30.0}=34.8\: \rm m/s$$