A test rocket is launched by accelerating it along an inclined ramp of length $L$ at acceleration $a$, starting from rest at point $A$. The ramp is positioned at an angle $\theta $ above the horizontal. At the instant the rocket leaves the ramp, its engines turn off and it is only under the influence of gravity. You can ignore any air resistance.

1. Find an expression for the velocity of the rocket at the instant it leaves the ramp in terms of $a$ and $L$.

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To find the speed of the rocket at the top of the ramp given that it accelerates with acceleration $a$ over a distance $L$, we use the formula $v^2_{fx}=v^2_{0x}+2a_x\mathrm{\Delta }x$ as follows

\begin{aligned} v^2_{fx}=v^2_{0x}+2a_x\mathrm{\Delta }x\ \ \ \ &\Rightarrow \ \ \ \ \ v^2_{fx}=0+2aL \\ \\ &\Rightarrow \ \ \ \ \ v_{fx}=\sqrt{2aL} \end{aligned}

The speed of the rocket as it launches off the ramp is therefore equal to

\boxed{v_{fx}=\sqrt{2aL}}

2. Find an expression for the maximum height above the ground reached by the rocket in terms of $a$, $L$, and $\theta $.

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The rocket launches off the ramp with a speed $v_0=\sqrt{2aL}$ and therefore the components of its initial velocity for the projectile motion part of its flight are

\begin{aligned} &v_{0x}=\sqrt{2aL}{\mathrm{cos} \left(\theta \right)\ } \\ &v_{0y}=\sqrt{2aL}{\mathrm{sin} \left(\theta \right)\ } \end{aligned}

and the kinematic equations describing its flight are

\left\{ \begin{array}{c} x\left(t\right)=\sqrt{2aL}{\mathrm{cos} \left(\theta \right)\ }t \\ \\ v_x\left(t\right)=\sqrt{2aL}{\mathrm{cos} \left(\theta \right)\ } \end{array} \right.\ \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c} \displaystyle{y\left(t\right)=-\frac{1}{2}gt^2+\sqrt{2aL}{\mathrm{sin} \left(\theta \right)\ }t+H} \\ \\ v_y\left(t\right)=-gt+\sqrt{2aL}{\mathrm{sin} \left(\theta \right)\ } \end{array} \right.

where $H=L{\mathrm{sin} \left(\theta \right)\ }$ is the height of the ramp.

The rocket reaches its maximum height at a time $t_1$ such that $v_y\left(t_1\right)=0$ and we solve for $t_1$ as follows

\begin{aligned} v_y\left(t_1\right)=0\ \ \ \ \ &\Rightarrow \ \ \ \ -gt_1+\sqrt{2aL}{\mathrm{sin} \left(\theta \right)\ }=0 \\ \\ &\Rightarrow \ \ \ \ \ t_1=\frac{\sqrt{2aL}}{g}{\mathrm{sin} \left(\theta \right)\ } \end{aligned}

\begin{aligned} H_{max}&=y\left(t_1\right) \\ \\ &=-\frac{1}{2}gt^2_1+\sqrt{2aL}{\mathrm{sin} \left(\theta \right)\ }t_1+L{\mathrm{sin} \left(\theta \right)\ } \\ \\ &=-\frac{1}{2}g\cdot {\left(\frac{\sqrt{2aL}}{g}{\mathrm{sin} \left(\theta \right)\ }\right)}^2+\sqrt{2aL}{\mathrm{sin} \left(\theta \right)\ }\cdot \frac{\sqrt{2aL}}{g}+L{\mathrm{sin} \left(\theta \right)\ } \\ \\ &=-\frac{1}{2}g\cdot \frac{2aL{{\mathrm{sin}}^{\mathrm{2}} \left(\theta \right)\ }}{g^2}+\frac{2aL{{\mathrm{sin}}^{\mathrm{2}} \left(\theta \right)\ }}{g}+L{\mathrm{sin} \left(\theta \right)\ } \\ \\ &=\frac{aL{{\mathrm{sin}}^2 \left(\theta \right)\ }}{g}+L{\mathrm{sin} \left(\theta \right)\ } \end{aligned}

The max height reached by the rocket is therefore equal to

\boxed{H_{max}=\frac{aL{{\mathrm{sin}}^2 \left(\theta \right)\ }}{g}+L{\mathrm{sin} \left(\theta \right)\ }}

3. Let $L=200\ m$, $a=1.25\ m/s^2$, and $\theta =35{}^\circ $. Find how far along the ground from point $A$ the rocket lands.

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We first derive the time of flight $t_f$ of the rocket, and then use it to compute the horizontal distance it travels while in flight. Eventually, we add this distance to the base of the incline which has length $L{\mathrm{cos} \left(\theta \right)\ }$ to derive the total distance traveled horizontally.

To find the time of flight $t_f$, we solve $y\left(t_f\right)=0$ as follows

\begin{aligned} y\left(t_f\right)=0\ \ \ \ &\Rightarrow \ \ \ \ \ -\frac{1}{2}gt^2_f+\sqrt{2aL}{\mathrm{sin} \left(\theta \right)\ }t_f+L{\mathrm{sin} \left(\theta \right)\ }=0 \\ \\ &\Rightarrow \ \ \ \ \ \frac{1}{2}gt^2_f-\sqrt{2aL}{\mathrm{sin} \left(\theta \right)\ }t_f-L{\mathrm{sin} \left(\theta \right)\ }=0 \\ \\ &\Rightarrow \ \ \ \ \left\{ \begin{array}{c} \displaystyle{t_f=\frac{\sqrt{2aL}{\mathrm{sin} \left(\theta \right)\ }-\sqrt{2aL{{\mathrm{sin}}^2 \left(\theta \right)+2gL{\mathrm{sin} \left(\theta \right)\ }\ }}}{g}} \\ \\ or \\ \\ \displaystyle{t_f=\frac{\sqrt{2aL}{\mathrm{sin} \left(\theta \right)\ }+\sqrt{2aL{{\mathrm{sin}}^2 \left(\theta \right)+2gL{\mathrm{sin} \left(\theta \right)\ }\ }}}{g}} \end{array} \right. \end{aligned}

Ignoring the first root which is negative, we conclude that the time of flight of the rocket is equal to

\begin{aligned} t_f&=\frac{\sqrt{2aL}{\mathrm{sin} \left(\theta \right)\ }+\sqrt{2aL{{\mathrm{sin}}^2 \left(\theta \right)-2gL{\mathrm{sin} \left(\theta \right)\ }\ }}}{g} \\ \\ &=\frac{\sqrt{2\cdot 1.35\cdot 200}{\mathrm{sin} \left(35\right)\ }+\sqrt{2\cdot 1.35\cdot 200{{\mathrm{sin}}^2 \left(35\right)+2\cdot 9.8\cdot 200{\mathrm{sin} \left(35\right)\ }\ }}}{9.8} \\ \\ &\approx 6.38\ s \end{aligned}

Thus, the distance from the right end of the ramp to the landing point is equal to

x\left(t_f\right)=\sqrt{2aL}{\mathrm{cos} \left(\theta \right)\ }t_f=\sqrt{2\cdot 1.35\cdot 200}{\mathrm{cos} \left(35\right)\ }\cdot 6.38\approx 121.44\ m

Finally, the distance between point $A$ and the landing point is equal to

\boxed{D=L{\mathrm{cos} \left(\theta \right)\ }+x\left(t_f\right)=200\cdot {\mathrm{cos} \left(35\right)\ }+121.44\approx 285.27\ m}

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