Question

A man who can throw a stone with a velocity of 20 m/s wishes to hit a target placed on his own level at a distance of 30m. At what angle should be threw the stone?

Answer 1

$$v_x(t)=v_0\cos \alpha$$

$$x(t)=x_0+v_0\cos \alpha \cdot t$$

$$v_y(t)=v_0\sin \alpha-gt$$

$$y(t)=y_0+v_0\sin \alpha\cdot t-\dfrac{gt^2}{2}$$

Let $x_0=x(0)=0, y_0=y(0)=0$

Find the time when $y=0$

$$0+v_0\sin \alpha\cdot t-\dfrac{gt^2}{2}=0$$

$$t_1=0, t_2=\dfrac{2v_0\cdot \sin \alpha}{g}$$

Then

$$x(t_2)=0+v_0\cos \alpha \cdot (\dfrac{2v_0\cdot \sin \alpha}{g})=30\ m$$

$$\sin 2\alpha=\dfrac{30\ m\cdot g}{2v_0^2}$$

$v_0=20\ m/s, g=9.81\ m/s^2$

$$\sin 2\alpha=\dfrac{30\ m\cdot 9.81\ m/s^2}{2(20\ m/s)^2}=0.367875$$

Since $0\degree\leq\alpha \leq90\degree$

$$2\alpha=\sin^{-1}(0.367875)$$

or

$$2\alpha=180\degree-\sin^{-1}(0.367875)$$

$$2\alpha\approx21.5846\degree \text{ or }2\alpha\approx158.4154\degree$$

$$\alpha\approx10.8\degree \text{ or }\alpha\approx79.2\degree$$