Question

A $500g$model rocket is resting horizontally at the top edge of a $40-m$-high wall when it is accidentally bumped. The bump pushes it off the edge with a horizontal speed of $0.5$m/s and at the same time causes the engine to ignite. When the engine fires, it exerts a constant $20N$horizontal thrust away from the wall.

a. How far from the base of the wall does the rocket land?

b. Describe the rocket’s trajectory as it travels to the ground

Short answer

a). The rocket will land $164\mathrm{m}$away from the base of the wall.

b). The trajectory is a straight line.

Step-by-step solution

Given Information (Part a)

Given in the question that,

$m$is the mass of the rocket model$=500g$

$h$is the height of the wall $=40m$

$F$is the force $=20N$

$u$is the initial horizontal velocity $=0.5m/s$

$g$is the vertical acceleration$=9.8m/s^2$

Explanation (Part a)

Let's apply the second law of motion to find $t$:

$h=ut+\frac{1}{2}g{t}^2$

$40=0+\frac{1}{2}\times 9.8\times t^2$

$t=2.86second$

We need to find the horizontal acceleration $a$:

Therefore,

$a=\frac{F}{m}$

$a=\frac{20}{0.5}$

$a=40m/s^2$

Let's apply the second law of motion again:

$s=ut+\frac{1}{2}a{t}^2$

where, $s$is the distance travelled time $t$

$s=0.5\times 2.86+\frac{1}{2}\times 40\times 2.{86}^2$

$s=164.7m$

Final Answer (Part a)

The rocket will land $164.7m$away from the base of the ball.

Given Information (Part b)

The mass of the rocket is $m=500\mathrm{g}=0.5\mathrm{kg}$, the initial velocity of the rocket is $v_i=0.5\mathrm{m}/\mathrm{s}$, the thrust of the rocket is$F_t=20\mathrm{N}$, the height of the rocket is $h=40\mathrm{m}$, and the gravitational acceleration is $g=9.8\mathrm{m}$.

Explanation (Part b)

Let's find the trajectory of the rocket by using given equation:

$x=v_{ix}t+\frac{1}{2}{a}_x{t}^2$

$=(0.5\mathrm{m}/\mathrm{s})t+\frac{1}{2}\left(40\mathrm{m}/{\mathrm{s}}^2\right)t^2$ (1)

Ignore the first term $(0.5\mathrm{m}/\mathrm{s})t$and rewrite the formula as follow:

$x=\left(20\mathrm{m}/{\mathrm{s}}^2\right)t^2$

Let's solve for $t^2$

$t^2=\frac{x}{\left(20\mathrm{m}/{\mathrm{s}}^2\right)}$ (2)

Let's consider the $y$position of the rocket:

$y=y_0-\frac{1}{2}{a}_y{t}^2$

$=\left(40\mathrm{m}\right)-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)t^2$ (3)

Let's substitute $t$from the equation (2) and (3):

$y=\left(40\mathrm{m}\right)-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\frac{x}{\left(20\mathrm{m}/{\mathrm{s}}^2\right)}$

$y=\left(40\mathrm{m}\right)-0.245x$

The equation shows that the trajectory is a straight line :

Final Answer (Part b)

The trajectory of the rocket is a straight line