Question

A test rocket is
launched by accelerating
it along a 200.0-m incline
at 1.25 m/s2 starting from
rest at point .The incline rises at
35.0 above the horizontal,
and at the instant the
rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored). Find (a) the maximum height above the ground that the rocket reaches, and (b) the greatest horizontal range of the rocket beyond point A.

Answer 1

Answer on Question#82201 - Physics - Mechanics | Relativity

A test rocket is launched by accelerating it along a 200.0-m incline at $1.25\mathrm{m} / \mathrm{s}^2$ starting from rest at point A. The incline rises at $35{}^{\circ}$ above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored). Find (a) the maximum height above the ground that the rocket reaches, and (b) the greatest horizontal range of the rocket beyond point A.

Solution:

Since $AB = 200 \, \text{m}$ , thus we obtain

$$B C = A B \sin 3 5 {}^ {\circ} = 1 1 4. 7 \mathrm {m}$$ $$A C = A B \cos 3 5 {}^ {\circ} = 1 6 3. 8 \mathrm {m}$$

The a speed $v_{f}$ , an initial speed $v_{i}$ , a traveled path $l$ and an acceleration $a$ are related by the following expression

$$v _ {f} ^ {2} - v _ {i} ^ {2} = 2 a l$$

Since $v_{i} = 0 \, \text{m/s}$ , $l = 200 \, \text{m}$ and $a = 1.25 \, \text{m/s}^2$ , we obtain the speed at point B:

$$v _ {f} = \sqrt {2 a l + v _ {i} ^ {2}} = \sqrt {2 \cdot 1 . 2 5 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} \cdot 2 0 0 \mathrm {m} + \left(0 \frac {\mathrm {m}}{\mathrm {s}}\right) ^ {2}} = 2 2. 3 6 \frac {\mathrm {m}}{\mathrm {s}}$$

The coordinates of the rocket after it passes point B is described by following system of equations (origin is at point A):

$$\left\{ \begin{array}{l} y (t) = B C + v _ {f} \sin 3 5 {}^ {\circ} t - \frac {g t ^ {2}}{2}, \\ x (t) = A C + v _ {f} \cos 3 5 {}^ {\circ} t \end{array} \right.$$

where $g = 9.8 \, \text{m/s}^2$ – is the acceleration due to gravity and $t$ – is time.

The vertical component of velocity is given by

$$v _ {y} = v _ {f} \sin 3 5 {}^ {\circ} - g t$$

The maximum height is reached when $v_{y} = 0$ , thus we obtain

$$t = \frac {v _ {f} \sin 3 5 {}^ {\circ}}{g} = \frac {2 2 . 3 6 \frac {\mathrm {m}}{\mathrm {s}} \cdot \sin 3 5 {}^ {\circ}}{9 . 8 \frac {\mathrm {m}}{\mathrm {s} ^ {2}}} = 1. 3 1 \mathrm {s}$$

Substituting it into $y(t)$ we get

$$y_{max} = y(1.31 \text{ s}) = 114.7 \text{ m} + 22.36 \frac{\text{m}}{\text{s}} \cdot \sin 35{}^\circ \cdot 1.31 \text{ s} - \frac{9.8 \frac{\text{m}}{\text{s}^2} \cdot (1.31 \text{ s})^2}{2}$$ $$y_{max} = 123.1 \text{ m}$$

The rocket will fall when $y(t) = 0$, therefore

$$BC + v_f \sin 35{}^\circ t - \frac{g t^2}{2} = 0$$ $$114.7 \text{ m} + 12.8 \frac{\text{m}}{\text{s}} \cdot t - \frac{9.8 \frac{\text{m}}{\text{s}^2} \cdot t^2}{2} = 0$$

The positive root of this equation is $t = 6.3$ s. It is the time when the rocket falls on the ground after it passes point B. Substituting it into $x(t)$ we obtain the greatest horizontal range:

$$x_{max} = x(6.3 \text{ s}) = 163.8 \text{ m} + 18.3 \frac{\text{m}}{\text{s}} \cdot 6.3 \text{ s} = 279.1 \text{ m}$$

Answer: $y_{max} = 123.1 \text{ m}, x_{max} = 279.1 \text{ m}$.

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