Question

A 600 g model rocket is on a cart that is rolling to the right at a speed of 2.5 m/s. The rocket engine, when it is fired, exerts a 8.0 N vertical thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is 20 m above the launch point.
At what horizontal distance left of the loop should you launch?

Answer 1

Answer: 8.43 m

Explanation:

Given

Mass of the rocket, m = 600 g = 0.6 kg

Speed of riling, v = 2.5 m/s

F = mg, where

F = weight of the rocket in N

m = mass of the rocket in kg

g = acceleration due to gravity in m/s²

F = 0.6 * 9.8

F = 5.88 N

Since the the vertical thrust on the rocket is 8 N and 5.88 N of it will be used to counter the rockets weight. Thus, the remaining 2.12 N is what is available to provide acceleration.

Upward acceleration of the rocket is..... F = ma, a = F / m

a = 2.12 / 0.6

a = 3.53 m/s²

using equation of motion, we will calculate how long it will take to rise 20 m in the air

S = ut + 1/2at², where u = 0

20 = 1/2 * 3.53 * t²

t² = 4. / 3.53

t² = 11.33

t = √11.33 = 3.37 s

The distance then is

v = Δx / t, such that

Δx = v * t

Δx = 2.5 * 3.37

Δx = 8.43 m

Answer 2

Answer:

8.66 m

Explanation:

600g = 0.6 kg

Let g = 10m/s2. Gravity acting on the rocket model would have a magnitude of:

mg = 0.6*10 = 6 N

When the thrust of 8N acting on the rocket, the net force would be F = 8 - 6 = 2N vertically upward. So the net acceleration according to Newton 2nd law would be

a = F/m = 2 / 0.6 = 3.33 m/s2

We can calculate the time t it would take to reach h = 20m vertically at this acceleration:

$h = at^2/2$

$t^2 = 2h/a = 2*20/3.33 = 12$

$t = \sqrt{12} = 3.46 s$

Assume that the horizontal speed v = 2.5m/s stays constant, then the horizontal distance from where it starts thrusting up to the loop is

2.5 * 3.46 = 8.66 m