Question

A $500g$ model rocket is on a cart that is rolling to the right at a speed of $3.0m/s$. The rocket engine, when it is fired, exerts an$8.0N$ vertical thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is $20m$ above the ground. At what horizontal distance left of the hoop should you launch?

Short answer

The horizontal distance between the hoop and the rocket should be$7.6\mathrm{m}$.

Step-by-step solution

Given Information

A $500g$ model rocket is on a cart that is rolling to the right at a speed of $3.0m/s$,

Explanation

In this problem, the car moving horizontally with a speed, and while the car is moving, the rocket is launched vertically, which means the rocket will move with two components, in$y$direction and in the $x$direction where the rocket gains its $x$component due to the movement of the car. The diagram below shows the situation. Where the rocket travels distances $x_f$and $y_f$where our target is to find $x_f$that the rocket passes through the hoop.

First, let us calculate the time taken by the rocket to travel vertically $t_{f,y}$. To find the time we should find the acceleration that the rocket gains due to the launch. The rocket is launched vertically due to its force $F_y=8\mathrm{N}$against its weight. So, as the net force on the rocket will be given by

$F_{\text{net}}=F_y-mg$

$m{a}_y=F_y-mg$

$a_y=\frac{F_y-mg}{m}$

Explanation

Now, we plug the values for $F_y,m$and $g$into equation (1) to get $a_y$

$a_y=\frac{F_y-mg}{m}$

$=\frac{8\mathrm{N}-(0.5\mathrm{kg})\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{0.5\mathrm{kg}}$

$=6.2\mathrm{m}/{\mathrm{s}}^2$

The rocket starts from an initial position $y_i=0$with an initial speed $v_{i,y}=0$and time $t_{i,y}=0$and reaches after travelling distance$y_f$in time $t_{f,y}$. We need to find the time $t_{f,y}$by using the value of the calculated $a_y$. The hoop is $20\mathrm{m}$above the launch point, so the distance that the rocket travel is $y_f=20\mathrm{m}$. From the kinematics equation, we can find the time $t_{f,y}$by

$y_f=y_i+{\left(v_i\right)}_y\left(t_{f,y}-t_{i,y}\right)+\frac{1}{2}{a}_y{\left(t_{f,y}-t_{i,y}\right)}^2$

Explanation

Now, we plug the values for $y_f,y_i,v_{i,y},t_{i,y}$and $a_y$into equation (2) to get $t_{f,y}$

$y_f=y_i+{\left(v_i\right)}_y\left(t_{f,y}-t_{i,y}\right)+\frac{1}{2}{a}_y{\left(t_{f,y}-t_{i,y}\right)}^2$

$20\mathrm{m}=0+0\left(t_{f,y}-t_{i,y}\right)+\frac{1}{2}\left(6.2\mathrm{m}/{\mathrm{s}}^2\right){\left(t_{f,y}-0\right)}^2$

$t_{f,y}=2.54\mathrm{s}$

This is the time that taken for the rocket to pass from the hoop. so the rocket needs to travel horizontally in time $t_{f,x}=t_{f,y}=2.5\mathrm{s}$

Explanation

For the horizontal component, the rocket starts from an initial position is $x_i=0$with an initial speed $v_i=3m/s$and reaches to the hoop after travelling distance $x_f$in time $t_f$. The rocket moves with constant speed, which means its acceleration equals zero . From the kinematics equation, we can find the distance $x_f$by

$x_f=x_i+{\left(v_i\right)}_x\left(t_f-t_i\right)+\frac{1}{2}{a}_x{\left(t_f-t_i\right)}^2$

Where $x_f$is the horizontal distance between the hoop and the rocket to be launched. Now, we plug the values for $x_i,v_i,t_i,t_f$and $a_x$into equation (3) to get $x_f$

$x_f=x_i+{\left(v_i\right)}_x\left(t_f-t_i\right)+\frac{1}{2}{a}_x{\left(t_f-t_i\right)}^2$

$=0+(3\mathrm{m}/\mathrm{s}{)}_x(2.59\mathrm{s}-0)+\frac{1}{2}(0){\left(t_{f,\mathrm{T}}-t_{i,\mathrm{T}}\right)}^2$

$=7.6m$

Final Answer

The horizontal distance between the hoop and the rocket should be $7.6\mathrm{m}$.