Project 1

Consider the general case for projectile motion from the origin.

In the horizontal direction we have $s=x$ , $u=v\cos\theta$ , and $a=0$ .

In the vertical direction we have $s=0$ , $u=v\sin\theta$ , and $a=-g$ (neglecting the height of the boy).

Using $s=ut+\frac{1}{2}at^2$ we obtain:

$x=vt\cos\theta+0\Rightarrow t=\frac{x}{v\cos\theta}$

and

$0=vt\sin\theta-\frac{1}{2}gt^2 \Rightarrow t=\frac{2v\sin\theta}{g}|t\neq0$

Equating these we have: $\frac{2v\sin\theta}{g}=\frac{x}{v\cos\theta}\Rightarrow x=\frac{v^2 \sin(2\theta)}{g}$

$\sin(2\times30)=\sin(90-30)$ and $\sin(2\times60)=\sin(90+30)$ which are equal by the local symmetry of sin about $90$ .

So for constant velocity the distance will be the same in both cases and it falls in the bucket again.

The horizontal distance that the ball covers is same for $\theta$ and $90-\theta$ .