Consider a box of mass $m$ being pushed along a frictionless horizontal surface which ends with a dropoff into a ravine of width $w$. You desire to push the box fast enough for it to fall across the ravine and land on the other side which is lower by a distance $h$.

1. Find the minimum velocity required for the box to just barely make it to the other side in terms of $w$, $h$ and any necessary constants.

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The box is launched horizontally with a velocity $v_0$ such that it barely lands on the other side of the ravine, as shown in the figure above. The kinematic equations of the box in flight along the $x$ and $y$ directions are given by

\left\{ \begin{array}{c} x\left(t\right)=v_0t \\ v_x\left(t\right)=v_0 \end{array} \right.\ \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c} \displaystyle{y\left(t\right)=-\frac{1}{2}gt^2+h} \\ \\ v_y\left(t\right)=-gt \end{array} \right.

We let $t_f$ denote the time of flight and therefore conclude that $\displaystyle{\left\{ \begin{array}{c}
x\left(t_f\right)=w \\
y\left(t_f\right)=0 \end{array}
\right.}$ which allows us first to solve for $t_f$ as follows

\begin{aligned} x\left(t_f\right)=w\ \ \ \ &\Rightarrow \ \ \ \ \ v_0t_f=w \\ \\ &\Rightarrow \ \ \ \ \ t_f=\frac{w}{v_0} \end{aligned}

and then solve $y\left(t_f\right)=0$ to derive $v_0$

\begin{aligned} y\left(t_f\right)=0\ \ \ \ &\Rightarrow \ \ \ \ \ -\frac{1}{2}gt^2_f+h=0 \\ \\ &\Rightarrow \ \ \ \ \ -\frac{1}{2}g\cdot {\left(\frac{w}{v_0}\right)}^2+h=0 \\ \\ &\Rightarrow \ \ \ \ \ \frac{gw^2}{2v^2_0}=h \\ \\ &\Rightarrow \ \ \ \ \ v_0=\sqrt{\frac{gw^2}{2h}} \end{aligned}

Thus, the minimum velocity required for the box to just barely make it to the other side is equal to

\boxed{v_0=\sqrt{\frac{gw^2}{2h}}}

2. Find the $x$ and $y$ components of the velocity of the box just before it hits the ground on the other side in terms of $w$, $h$, and any necessary constants.

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The components of the velocity when the box hits the ground are equal to

\boxed{ \begin{aligned} v_x\left(t_f\right)&=v_0=\sqrt{\frac{gw^2}{2h}} \\ \\ v_y\left(t_f\right)&=-gt_f=-g\cdot \frac{w}{v_0}=-gw\ \sqrt{\frac{2h}{gw^2}}=-\sqrt{2gh} \end{aligned} }

3. Find the work done by gravity on the box as it crosses the ravine in terms of $m$, $h$, and any necessary constants.

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The work done by gravity on the box as it falls from $h_i=h$ to $h_f=0$ is equal to

\boxed{W_{mg}=-\mathrm{\Delta }U_g=-\left(mgh_f-mgh_i\right)=-\left(0-mgh\right)=mgh}

4. If the box was pushed with a horizontal force $F$, how far would it have to be pushed to reach the minimum velocity from part 1., in terms of $w$, $h$, $F$, and any necessary constants?

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Let $D$ denote the distance that the box has to be pushed over to acquire speed $v_0$ from rest. We first derive the acceleration of the box in the $x$-direction by writing Newton’s Second Law.

\begin{aligned} F_{net\ x}=ma_x\ \ \ \ &\Rightarrow \ \ \ \ \ F=ma_x \\ \\ &\Rightarrow \ \ \ \ \ a_x=\frac{F}{m} \end{aligned}

We then use the kinematic formula $v^2_{fx}=v^2_{0x}+2a_x\mathrm{\Delta }x$ to derive the distance $D$ as follows

\begin{aligned} v^2_{fx}=v^2_0+2a_x\mathrm{\Delta }x\ \ \ \ &\Rightarrow \ \ \ \ \ v^2_0=0+2\cdot \frac{F}{m}D \\ \\ &\Rightarrow \ \ \ \ \ D=\frac{mv^2_0}{2F} \\ \\ &\Rightarrow \ \ \ \ \ D=\frac{mgw^2}{2Fh} \end{aligned}

Thus, the minimum distance over which the box has to be pushed with force $F$ to reach a launch speed of $v_0$ is equal to

\boxed{D=\frac{mgw^2}{2Fh}}

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