Question

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 6600 m. You can ignore any effects of air resistance.

Required:
a. What was the rocket's acceleration during the first 16s?
b. What is the rocket's speed as it passes through acloud 5100 m above the ground?

Answer 1

a) a = 34.375 m / s², b) v_f = 550 m / s

Step-by-step explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

v_f =
`(x-x_1)/(t)`

we substitute the values

v_f =
`( 6600 -x_1)/(4)`

The initial part of the movement is carried out with acceleration

v_f = v₀ + a t

x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

x₁ = ½ a t²

v_f = a t

we substitute the values

x₁ = 1/2 a 16²

x₁ = 128 a

v_f = 16 a

let's write our system of equations

v_f =
`(6600 - x_1)/(4)`

x₁ = 128 a

v_f = 16 a

we substitute in the first equation

16 a =
`(6600 -128 a)/(4)`

16 4 a = 6600 - 128 a

a (64 + 128) = 6600

a = 6600/192

a = 34.375 m / s²

b) let's find the time to reach this height

x = ½ to t²

t² = 2y / a

t² = 2 5100 / 34.375

t² = 296.72

t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

v_f = 16 a

v_f = 16 34.375

v_f = 550 m / s