Question

A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s . The rocket engine, when it is fired, exerts a 7.0 N vertical thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is 15 m above the launch point.At what horizontal distance left of the hoop should you launch?

Answer 1

Answer:

The rocket has to be launched 8 m from the hoop

Explanation:

Let's analyze this problem, the rocket is on a car that moves horizontally, so the rocket also has the same speed as the car; The initial horizontal rocket speed is (v₀ₓ = 3.0 m/s).

On the other hand, when starting the engines we have a vertical force, which creates an acceleration in the vertical axis, let's use Newton's second law to find this vertical acceleration

    F -W = m a

    a = (F-mg) / m

    a = F/m  -g

    a = 7.0/0.500  - 9.8

    a = 4.2 m/s²

We see that we have a positive acceleration and that is what we are going to use in the parabolic motion equations

Let's look for the time it takes for the rocket to reach the height (y = 15m) of the hoop, when the rocket fires its initial vertical velocity is zero (I'm going = 0)

    y = $v_{oy}$ t + ½ a t²

    y = 0 + ½ a t²

    t = √ 2y/a

    t = √( 2 15 / 4.2)

    t = 2.67 s

This time is also the one that takes in the horizontal movement, let's calculate how far it travels

    x = v₀ₓ t

    x = 3 2.67

    x = 8 m

The rocket has to be launched 8 m from the hoop