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A basketball is launched by a person with an initial velocity $v$ at an angle $\theta $ from a height $L$ into a basket of height $H>L$ which is a horizontal distance $d$ from the person.

Additionally, $H-L=h$

Assumptions:

Zero air resistance and the diameter of the ball $<<$ diameter of the basket.

I am trying to prove:

$$v\sin(\theta) > \sqrt{2gh}$$

What I have done so far:

$$a_x=\ddot{x}=0$$

$$v_x=\dot{x}(t)=v\cos{\theta}$$

$$x(t)=vt\cos{\theta}$$ $$a_y=\ddot{y}=-g$$

$$v_y=\dot{y}(t)=L-gt$$

$$y(t)=L+v\sin(\theta)t-\frac{1}{2}gt^2$$

Assume the ball reaches the basket at time $t=t_b$

$$x(t_b)=vt_b\cos{\theta}=d \Longrightarrow t_b=\frac{d}{v\cos{\theta}}$$

$$y(t_b)=L+v\sin(\theta)t_b-\frac{1}{2}gt_b^2=H$$

$$H-L=v\sin(\theta)t_b-\frac{1}{2}gt_b^2=h$$

How do I proceed from here?

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I always start with a picture:

enter image description here

Horizontal velocity is $v\cos\theta$, so time to reach basket is $\frac{v\cos\theta}{d}$

We need to increase our height by at least $h$ in that time. It is intuitively obvious that the lowest velocity I can get away with is the one that will just get into the basket - that is, the ball arrives "horizontally". For the ball to follow the trajectory I drew, with a maximum along the way, it must have been going faster - more vertical velocity to reach a greater height.

If the ball just reaches $h$, then from conservation of energy we know $\frac12 m v_v^2 = m g h$, so the vertical component of velocity must be $v_v=\sqrt{2gh}$.

And the vertical component of velocity is $v_v = v\sin\theta$, so your expression immediately follows.

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If the Ball Never Goes Above $H$

In this case, all of the ball's initial kinetic energy (due to vertical motion) is converted into potential energy by the time it rises by distance $h$. Then, we have:

\begin{equation} \begin{split} \Delta K &= \Delta P \\ \frac{1}{2}m(v\sin{\theta})^2 &= mgh \\ v\sin{\theta} &= \sqrt{2gh} \end{split} \end{equation}

Where $\Delta K$ is the kinetic energy lost to the potential energy gained over distance $h$ and $\Delta P$ is the potential energy gained over the distance $h$.

If the Ball Goes Above $H$

The ball must go higher than $H$ to enter the basket.

In this case, not all of the ball's initial kinetic energy (due to vertical motion) is converted to potential energy by the time the ball has traveled over distance $h$. Thus $\frac{1}{2}m(v\sin{\theta})^2$ accounts for additional kinetic energy besides that which equals $mgh$.

So, we let

\begin{equation} \begin{split} K &= K_h + K_a &= \frac{1}{2}m(v\sin{\theta})^2 \\ P &= P_h + P_a &= (mgh) + P_a \\ K_h &= P_h \\ K_a &= P_a \end{split} \end{equation}

where:

  • $K$ is the total kinetic energy the ball loses while going up, due to changes in the vertical component of its velocity.
  • $P$ is the total potential energy the ball gains while going up.
  • Subscript $_h$ refers to quantities corresponding to motion from height $L$ to height $H$, over distance $h$.
  • Subscript $_a$ refers to quantities corresponding to motion at heights above $H$.

It follows that:

\begin{equation} \begin{split} K &= P \\ K - K_a &= P - P_a \\ K - K_a &= P_h \\ \frac{1}{2}m(v\sin{\theta})^2 - K_a &= mgh \\ (v\sin{\theta})^2 &= 2gh + \frac{2}{m}K_a\\ v\sin{\theta} &= \sqrt{2gh + \frac{2}{m}K_a} \end{split} \end{equation}

Since $K_a$ is non-zero in this case, we have:

\begin{equation} \begin{split} v\sin{\theta} &> \sqrt{2gh} \end{split} \end{equation}

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