Answer to Question #202114 in Physics for xyjcai

Question

 How will you explain the relationship between the angle of release to the 

 height of the projectile? How about angle of release and range before it 

 reaches the farthest distance?

Answer 1

The best way to explain the relationship it to derive it. So, the height can be found as

$$h=\frac{[v_y^2]}{2g}=\frac{[(v\sin\theta)^2]}{2g}=\frac{v^2\sin^2\theta}{2g}.$$

As we see,

$$h(\theta)=\frac{v^2}{2g}\sin^2\theta.$$

Find the range:

$$R=[v_x][t]=[v\cos\theta]\bigg[2\frac{[v_y]}{g}\bigg]=[v\cos\theta]\bigg[2\frac{[v\sin\theta]}{g}\bigg],\\\space\\ R(\theta)=\frac{2v^2\sin\theta\cos\theta}{g}=\frac{v^2\sin(2\theta)}{g}.$$

The height depends on the squared sine of angle of launch, the range depends on the sine of double angle of launch.