Projectile Motion - Speed Variations has no impact

Question

https://www.youtube.com/watch?v=0jGZnMf3rPo Explain why the bullet would always hit the monkey by solving the projectile problem.

Despite the initial speed to be of 13m/s or 9m/s, the monkey gets hit.

Answer 1

If the mouth of the cannon is at $(0,0)$ we have the following schematic:

The height of the monkey above the cannon is $H$ and it's horizontal distance from the cannon is $L$. The initial speed of the ball is $V_0$ at an angle of $\alpha$ degrees.

We can then write the following equations for the monkey's $x$ and $y$ components: $$y_m = H - \frac{1}{2}gt^2\tag{1}$$ $$x_m = L$$ And for the ball we have: $$y_b= (V_0\sin \alpha) t- \frac{1}{2}gt^2$$ $$x_b = (V_0\cos \alpha) t$$ But $\sin \alpha = \frac{H}{\sqrt{H^2+L^2}}$ and $\cos \alpha = \frac{L}{\sqrt{H^2+L^2}}$. Inserting these into the equations for the ball we get: $$y_b= (V_0\frac{H}{\sqrt{H^2+L^2}}) t- \frac{1}{2}gt^2$$ $$x_b = (V_0\frac{L}{\sqrt{H^2+L^2}}) t$$ Let us see at what time $t_1$ the ball is at the same $x$ coordinate at the monkey by setting $x_b=x_m$, i.e.: $$ (V_0\frac{L}{\sqrt{H^2+L^2}}) t_1 = L$$ or $$ t_1 = \frac{\sqrt{H^2+L^2}}{V_0}$$ Inserting $t_1$ into the equation for the $y$ component of the ball, gives the height of the ball at this time: $$y_b= (V_0\frac{H}{\sqrt{H^2+L^2}}) t_1- \frac{1}{2}gt_1^2$$ or $$y_b= H- \frac{1}{2}gt_1^2$$ which is exactly the height that the monkey would be at, at that time (see equation 1).