Question

A catapult launches a rocket at an angle of$\mathbf{53}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}$above the horizontal with an initial speed of$\mathbf{100}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. The rocket engine immediately starts a burn, and for$\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{s}}$the rocket moves along its initial line of motion with an acceleration of$\mathbf{30}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. Then its engine fails, and the rocket proceeds to move in free fall. Find

(a) the maximum altitude reached by the rocket,

(b) its total time of flight, and

(c) its horizontal range.

Short answer

(a) The maximum altitude reached by the rocket is$1.52\text{km}$.

(b) The total time of flight of the rocket is$36.1\text{s}$.

(c) The horizontal range of the rocket is$4.05\text{km}$.

Step-by-step solution

Given data

The initial speed of the rocket is$100\mathrm{m}/\mathrm{s}$with an angle$53.0^{\circ }$above the horizontal, the time for which the rocket runs along its initial line of motion is 33 and the acceleration is $30.0\mathrm{m}/{\mathrm{s}}^2$.

The formula to calculate the vertical height reached by the rocket.

The formula to calculate the vertical height reached by the rocket

$\mathit{h}\mathbf{=}\left({\left(v_0\right)}_{y}sin\theta \right)\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{a}}_{\mathbf{y}}{\mathit{t}}^{\mathbf{2}}$...... (1)

Here,

$\mathit{h}$is the vertical height, ${\mathbf{\left(}{\mathbf{v}}_{\mathbf{0}}\mathbf{\right)}}_{\mathbf{y}}$is the vertical component of the initial velocity, $\mathit{t}$is the time taken by the rocket,${\mathit{a}}_{\mathbf{y}}$is the vertical component acceleration of the rocket,$\mathit{\theta }$is the angle made by the rocket with the horizontal.

The speed of the rocket after the failure Is given by

${\mathbf{\text{v}}}_{\mathbf{0}}\mathbf{=}{\left(v_0\right)}_{\mathbf{y}}\mathbf{+}{\mathit{a}}_{\mathbf{y}}\mathit{t}$...... (2)

The height reached by the rocket after the engine failure is,

${\mathit{h}}_{\mathbf{f}}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{-}{\mathbf{v}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }}{\mathbf{-}\mathbf{2}\mathbf{g}}$....... (3)

Here,$\mathbf{v}$is the final velocity of the rocket,$\mathbf{g}$is the acceleration due to gravity,${\mathbf{h}}_{\mathbf{f}}$is the height reached by the rocket after the engine failure.

The horizontal range of the rocket during constant acceleration is,

${\mathit{R}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{h}}^{\mathbf{\text{'}}}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$

Here,${\mathit{R}}^{\mathbf{\text{'}}}$is the horizontal range of the rocket during constant acceleration.

Determine the maximum altitude reached by the rocket:

(a)

Substitute$100\mathrm{m}/\mathrm{s}$for${\left(v_0\right)}_{\mathrm{y}},53.0°$for$\theta ,30.0\mathrm{m}/{\mathrm{s}}^2$for$a_{\mathrm{y}}$and$3.00\mathrm{s}$for$t$in the

$$\begin{gathered} h=\left({\left(v_0\right)}_{\mathrm{y}}\sin \theta \right)t+\frac{1}{2}{a}_{\mathrm{y}}{t}^2 \\ \begin{array}{c}h=\left((100\mathrm{m}/\mathrm{s})\sin \left(53.0°\right)\right)(3.00\mathrm{s})+\frac{1}{2}\left(30.0\mathrm{m}/{\mathrm{s}}^2\right)(3.00\mathrm{s}{)}^2\\ =\left((100\mathrm{m}/\mathrm{s})\sin \left(53.0°\right)\right)(3.00\mathrm{s})+\frac{1}{2}\left(30.0\mathrm{m}/{\mathrm{s}}^2\right)(3.00\mathrm{s}{)}^2\\ =239.59\mathrm{m}+135\mathrm{m}\\ =374.59\mathrm{m}\end{array} \end{gathered}$$

Thus, the vertical height of rocket is $374.59\mathrm{m}$.

The speed of the rocket after the failure Is given by

$v_0={\left(v_0\right)}_y+a_{y}t$

Substitute $100\mathrm{m}/\mathrm{s}$for ${\left(v_0\right)}_{\mathrm{y}},30.0\mathrm{m}/{\mathrm{s}}^2$for $a_{\mathrm{y}}$and $3.00\mathrm{s}$for $t$in the above equation.

$\begin{array}{c}{v}_0=100\mathrm{m}/\mathrm{s}+\left(30.0\mathrm{m}/{\mathrm{s}}^2\right)(3.00\mathrm{s})\\ =190\mathrm{m}/\mathrm{s}\end{array}$

The height reached by the rocket after the engine failure is,

$h_f=\frac{v^2-v_0^{2}si{n}^2\theta }{-2g}$

Substitute 0 for$v,190\mathrm{m}/\mathrm{s}$for$v_0,53.0^{\circ }$for$\theta$and$10.0\mathrm{m}/{\mathrm{s}}^2$for$g$in the above equation,

$\begin{array}{c}{h}_{\mathrm{f}}=\frac{0^2-{(190\mathrm{m}/\mathrm{s})}^2{\sin }^2\left(53.0\right)}{-2\left(10.0\mathrm{m}/{\mathrm{s}}^2\right)}\\ =1151.26\mathrm{m}\end{array}$

The maximum height reached by the rocket is,

$h_{\max }=h_{\mathrm{f}}+h$

Substitute $1151.26\text{m}$for $h_f$and $374.59\text{m}$for $h$in the above equation.

role="math" localid="1663618849686" $\begin{array}{c}{h}_{\max }=1151.26\mathrm{m}+374.59\mathrm{m}\\ =1525.8\mathrm{m}\left(\frac{{10}^{-3}\mathrm{km}}{1\mathrm{m}}\right)\\ =1.52\mathrm{km}\end{array}$

Therefore, the maximum altitude reached by the rocket is $1.52\text{km}$.

The total time of flight of the rocket.

(b)

The vertical height reached by the rocket at the time of free fall is,

$h=\left(v_{0}sin\theta \right)t^{\text{'}}+\frac{1}{2}g{t}^{\text{'}2}$

Substitute $190\mathrm{m}/\mathrm{s}$for $v_0,53.0$for $\theta$,$9.8\mathrm{m}/{\mathrm{s}}^2$for $g$and $374.59\mathrm{m}$for $h$in the above equation.

$\begin{array}{c}-374.59\mathrm{m}=(190\mathrm{m}/\mathrm{s})\sin \left(53.0°\right)t^{\text{'}}+\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)t^{\text{'}2}\\ -374.59\mathrm{m}=151.7\mathrm{m}/\mathrm{s}{t}^{\text{'}}-\left(4.9\mathrm{m}/{\mathrm{s}}^2\right)t^{\text{'}2}\\ t^{\text{'}}=33.10\mathrm{s},-2.14\mathrm{s}\\ t^{\text{'}}=33.10\mathrm{s}\end{array}$

The total time of the flight of the rocket is,

$T=t+t\text{'}$

Substitute$33.10\text{s}$for$t\text{'}$and$3\text{s}$for$t$in the above equation.

$\begin{array}{c}T=3.00\mathrm{s}+33.10\mathrm{s}\\ =36.1\mathrm{s}\end{array}$

Therefore, the total time of flight of the rocket is $36.1\mathrm{s}$.

Determine the horizontal range of the rocket:

The formula to calculate the displacement of the rocket along the initial line of motion is,

$h^{\text{'}}={\left(v_0\right)}_{y}t+\frac{1}{2}{a}_y{t}^2$

Substitute $100\mathrm{m}/\mathrm{s}$for ${\left(v_0\right)}_{\mathrm{y}},30.0\mathrm{m}/{\mathrm{s}}^2$for $a_y$and $3\text{s}$for $t$in the above equation.

$\begin{array}{c}{h}^{\text{'}}=(100\mathrm{m}/\mathrm{s})(3.00\mathrm{s})+\frac{1}{2}\left(30.0\mathrm{m}/{\mathrm{s}}^2\right)(3.00\mathrm{s}{)}^2\\ =300.0\mathrm{m}+135.5\mathrm{m}\\ =435.0\mathrm{m}\end{array}$

Thus, the displacement of the rocket along the initial line of motion is $435.0\mathrm{m}$.

The horizontal range of the rocket during constant acceleration is,

$R^{\text{'}}=h^{\text{'}}cos\theta$

Substitute$435\text{m}$for$h$and${53}^{\circ }$for$\theta$in the above equation.

$\begin{array}{c}{R}^{\text{'}}=(435.0\mathrm{m})\cos \left(53.0\right)\\ =261.8\mathrm{m}\end{array}$

The horizontal range of the rocket during free fall is calculated as,

$R^{\text{'}\text{'}}=\left(v_0\cos \theta \right)T$

Substitute$190m/s$for$v_0,53.0$for$\theta$and$33.10\text{s}$for$T$in the above equation.

$\begin{array}{c}{R}^{\text{'}\text{'}}=(190\mathrm{m}/\mathrm{s})\cos \left(53.0°\right)(33.1\mathrm{s})\\ =3784.81\mathrm{m}\end{array}$

The total horizontal range of the rocket is,

$R=R^{\text{'}}+R^{\text{'}\text{'}}$

Substitute$261.8\text{m}$for$R\text{'}$and$3784.81\text{m}$for$R\text{'}\text{'}$in the above equation.

$\begin{array}{c}R=261.8\mathrm{m}+3784.81\mathrm{m}\\ =4046.61\mathrm{m}\left(\frac{{10}^{-3}\mathrm{km}}{1\mathrm{m}}\right)\\ =4.0466\mathrm{km}\\ \approx 4.05\mathrm{km}\end{array}$

Therefore, the horizontal range of the rocket is $4.05\mathrm{km}$.