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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 &= (mvh) + P_a \\ K_h &= P_h \\ K_a &= P_a \end{split} \end{equation}\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}

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 &= (mvh) + 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}

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}

Corrected an algebraic error.
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quietopus
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  • 21
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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 &= (mvh) + 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}-\sqrt{\frac{2}{m}K_a} &= \sqrt{2gh} \\ v\sin{\theta} &= \sqrt{2gh} + \sqrt{\frac{2}{m}K_a} \end{split} \end{equation}\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} + \sqrt{\frac{2}{m}K_a} \\ v\sin{\theta} &> \sqrt{2gh} \end{split} \end{equation}\begin{equation} \begin{split} v\sin{\theta} &> \sqrt{2gh} \end{split} \end{equation}

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 &= (mvh) + 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}-\sqrt{\frac{2}{m}K_a} &= \sqrt{2gh} \\ v\sin{\theta} &= \sqrt{2gh} + \sqrt{\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} + \sqrt{\frac{2}{m}K_a} \\ v\sin{\theta} &> \sqrt{2gh} \end{split} \end{equation}

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 &= (mvh) + 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}

edited body
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quietopus
quietopus
  • 21
  • 2

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

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}v\sin{\theta} \\ P &= P_h + P_a = mvh + 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}-\sqrt{\frac{2}{m}K_a} &= \sqrt{2gh} \\ v\sin{\theta} &= \sqrt{2gh} + \sqrt{\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} + \sqrt{\frac{2}{m}K_a} \\ v\sin{\theta} &> \sqrt{2gh} \end{split} \end{equation}

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 &= (mvh) + 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}-\sqrt{\frac{2}{m}K_a} &= \sqrt{2gh} \\ v\sin{\theta} &= \sqrt{2gh} + \sqrt{\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} + \sqrt{\frac{2}{m}K_a} \\ v\sin{\theta} &> \sqrt{2gh} \end{split} \end{equation}

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added 12 characters in body
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Edited to reflect a more correct reading of the problem statement.
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Clarified a statement about vertical motion.
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Clarified a statement about vertical motion.
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Revised some equations for clarity.
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Removed the longer derivation. Expanded and clarified explanation of problems in the question structure.
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Clarified a variable name.
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deleted 122 characters in body; added 2 characters in body
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