Please help: What horizontal distance left of the loop should you launch?

A 400g model rocket is on a cart that is rolling to the right at a speed of 2.5m/s . The rocket engine, when it is fired, exerts an 9.0N thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is 20m above the launch point. At what horizontal distance left of the loop should you launch?

Answer 1

F = ma
a = F/m = 9.0N/0.400kg = 22.5 m/s^2
Acceleration of the rocket is 22.5 m/s^2. Acceleration due to gravity is -9.81 m/s^2
Net acceleration: 22.5-9.81 = 12.69 m/s^2

Time to travel 20 m above launch point: y = y0+v0t+at^2/2
20 = 0+0t+(12.69t^2)/2
t = √[(2*20)/12.69] = 1.775 s

Horizontal velocity: Vx = 2.5 m/s.
Horizontal distance: x = Vx*t = 2.5*1.775 = 4.4 m

Answer 2

According to this information, I believe that initially the rocket is rolling on a horizontal surface. For the rocket to pass through a small horizontal hoop that is 20m above the launch point it must either be launched at a specific angle above horizontal, or it rolls into the bottom of hoop. Since we don’t know the horizontal distance to the left of the hoop, the rocket must roll into the bottom of the hoop. Then it moves from the bottom to the top of the hoop. To determine the horizontal distance to the left of the hoop, we need to determine the minimum velocity that required for the rocket to pass through the bottom of the hoop. To determine this number, we need to determine the minimum velocity that required for the rocket to pass through the top of the hoop.

When the rocket is at the top of the hoop, the centripetal force is equal its weight
Weight = 0.4 * 9.8 = 3.92 N
Fc = m * v^2/r
r is equal to one half of the height of the hoop.

Fc = 0.4 * v^2/10 = 0.04 * v^2
Set this equal to the weight and solve for v.

0.04 * v^2 = 3.92
v^2 = 3.92 ÷ 0.04 = 98
v = √98

This is approximately 9.9 m/s^2. Let’s use conservation of kinetic and potential energy to determine the rocket’s velocity as it enters the bottom of the loop.

KE = ½ * m * v^2 = ½ * 0.4 * 98 = 19.6 J
PE = m * g * h = 0.4 * 9.8 * 20 = 78.4 J
Total energy = 19.6 + 78.4 = 98 J

As the rocket moved from the bottom of the loop to the top of the loop, its kinetic energy was converted into the energy above. Let’s use the following equation to determine rocket’s velocity when it enters the bottom of the loop.

KE = ½ * 0.4 * v^2 = 0.2 * v^2
0.2 * v^2 = 98
v^2 = 98 ÷ 0.2 = 490
v = √490
This is approximately 22.1 m/s. Let’s use the following equation to determine the distance.

vf^2 = vi^2 + 2 * a * d
vf^2 = 490, vi^2 = 2.5^2 = 6.25
To determine the acceleration, use the following equation.

F = m * a
9 = 0.4 * a
a= 9/0.4 = 22.5 m/s^2

490 = 6.25 + 2 * 22.5 * d
45 * d = 490 – 6.25
d = 483.75 ÷ 45 = 10.75 meters
I hope this helps you to understand how to solve this type of problem.

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