Q43P - University Physics with Modern Physics

Question

A test rocket starting from rest at point A is launched by accelerating it along a 200.0 m incline at $\mathbf{1}\mathbf{.}\mathbf{90}\mathbf{ }\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$ (Fig. P3.43). The incline rises at  above the horizontal, and at the instant the rocket leaves it, the engines turn off and the rocket is subject to gravity only (ignore air resistance). Find (a) the maximum height above the ground that the rocket reaches, and (b) the rocket’s greatest horizontal range beyond point A.

Step-by-Step Solution

Answer

a) The maximum height of the rocket above the ground is 128m .

b) The maximum horizontal range of the rocket is 315 m

Step 1: Identification of given data

The given data can be listed below,

• The acceleration of the rocket is,$a=1.90 m/s^2$
• The angle of inclination is,$\theta =35°$

Step 2: Concept/Significance of horizontal range.

A projectile's horizontal range is the length of the horizontal plane it would traverse before returning to its initial vertical location.

Step 3: (a) Determination of the maximum height above the ground that the rocket

The initial height of the projectile launched is given by,

$y_0=d \sin \theta$

Here, d is the inclined distance.

Substitute all the values in the above,

$$\begin{gathered} y_0=\left(200m\right)\sin 35° \\ =114.7 m \end{gathered}$$

The vertical component of initial velocity is given by,

$v_y=v\sin \theta$

Here, v is the velocity after a displacement d.

Substitute all the values in the above,

$$\begin{gathered} v_{0y}=v\sin 35° \\ =\sqrt{2ad}\sin 35° \\ =\sqrt{2\left(1.90m/s^2\right)\left(200m\right)}\sin 35° \\ =15.81 m/s \end{gathered}$$

The equation for vertical motion of the projectile is given by,

$v_y^2=v_{0y}^2-2g\left(y-y_0\right)$

Here, $v_{0y}$ is initial vertical velocity of the projectile, g is the acceleration due to gravity, y is the maximum height of the projectile.

At maximum height, vertical velocity of the projectile is zero so the maximum height is given by,

$$\begin{gathered} 0=v_{0y}^2-2g\left(y_{max}-y_0\right) \\ {y}_{max}=y_0+\frac{v_{0y}^2}{2g} \end{gathered}$$

Substitute all the value in the above,

$$\begin{gathered} y_{max}=114.7 m+\frac{\left(15.81 m/s^2\right)}{2\left(9.8 m/s^2\right)} \\ =127.45 m \\ \approx 128 m \end{gathered}$$

Thus, the maximumheight of the rocket above the ground is 128 m .

Step 4: (b) Determination of the rocket’s greatest horizontal range beyond point A.

The time of flight of the rocket is given by,

$t=-\frac{1}{g}\left(-v_{0y}\pm \sqrt{v_{0y}^2+2g{y}_0}\right)$

Here,  $v_{0y}$ is initial vertical velocity of the projectile, g is the acceleration due to gravity, $y_0$ is the maximum height of the projectile.

Substitute all the values in the above,

$$\begin{gathered} t=-\frac{1}{9.8 m/s^2}\left[-15.81m/s\pm \sqrt{{\left(15.81m/s\right)}^2+2\left(9.8m/s^2\right)\left(114.7m\right)}\right] \\ =1.61s\pm 5.10s \end{gathered}$$

The positive value of time is taken to find the horizontal range which can be given by,

$x_{max}=d\cos \theta +t\sqrt{2ad}\cos \theta$

Here, d is the inclined distance, a is the acceleration, and t is the time of flight.

Substitute all the values in the above,

$$\begin{gathered} x_{max}=\left(200s\right)\cos 35°+\sqrt{2\left(1.90m/s^2\right)\left(200m\right)}\cos 35° \\ =315 m \end{gathered}$$

Thus, the maximum horizontal range of the rocket is 315 m