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?

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:

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

the focus of lesson 1 is newton's first law of motion - sometimes referred to as the law of inertia. newton's first law of motion is often stated as. an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

explanation:

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s