a rocket is launched from rest straight upwards at 20 m/s/s. after 10 seconds, the rocket engine stops

Complete Question: A rocket is launched from rest straight upwards at 20 m/s/s. After 10 seconds, the rocket engine stops. 1) What is the maximum height the rocket goes? 2) What is the speed of the rocket just before it hits the ground? 3) How long is the rocket in the air?

Answer:

1) 3,040 m. 2) 244.1 m/s 3) 55.3 sec.

Explanation:

1) We can split the height reached by the rocket in two parts:

1a) The rocket goes up at constant acceleration of 20 m/s², starting from rest.

As the acceleration is constant, we can use any of the kinematic equations in order to get the maximum height. As we know the time, the most suitable one is the following:

h₁ = 1/2*a*t² (as v₀=0) ⇒ h₁ = 1/2*20 m/s² * (10s) ² = 1,000 m

1b) In this part, the only acceleration (slowing down the rocket) is g. We need to know the value of v when the rocket engine stops, as this will be the initial velocity for this part.

We can find this value just applying the definition of acceleration, as follows:

vf = v₀ + a*t = 0 + 20 m/s² * 10 s = 200 m/s

Now, when the rocket reaches to the maximum height, it momentarily comes to an stop, so vf = 0.

We can apply this kinematic equation:

vf²-v₀² = 2*g*Δh. Solving for Δh:

Δh = (200m/s) ² / 2*9.8 m/s² = 2,040 m

So, the maximum height at which the rocket goes, is the sum of these two parts:

h = 1,000 m + 2,040 m = 3,040 m.

2) Once reached the height that we found out above, the rocket falls freely under the influence of gravity, starting from rest,

The speed of the rocket just before it hits ground, can be obtained from this equation, making v₀ = 0:

vf²-v₀² = 2*g*Δh ⇒vf = √2*g*Δh = √2*9.8m/s²*3,040m = 244.1 m/s

c) In order to get the total time that the rocket was in the air, we need to add the 3 different times, according what we have just done before:

t = t₁ + t₂ + t₃, where:

t₁ = time during which the rocket engine was on = 10 s

t₂ = time from the instant the rocket engine stops till the rocket reaches to the maximum height.

We can find this time, as we know the values of v₀, vf and a (g) for this part, as follows:

0 = v₀-g*t ⇒ t₂ = v₀ / g = 200 m/s / 9.8 m/s² = 20.4 s

t₃ = time during which the rocket fall freely from the maximum height to ground.

We can use this equation, solving for t, as v₀=0 (once reached to the maximum height, it comes momentarily to an stop, prior to change direction and fall freely), as follows:

h = 1/2*g*t² ⇒ t² = 2*h / g = 2*3,040 m / 9.8 m/s = 620.4 s²

⇒ t₃ = √620.4 s² = 24.9 s

⇒ t = t₁ + t₂ + t₃ = 10 s + 20.4 s + 24.9 s = 55.3 s