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 5100 m. you can ignore any effects of air resistance.
a. what was the rocket's acceleration during the first 16 s?
b. what is the rocket's speed as it passes through a cloud 5100 m above the ground?

Answer 1

consider the motion of rocket from t = 0 to t = 16 s

v₀ = velocity of rocket at t = 0 sec , = 0 m/s

Y₀ = position of rocket at t = 0 sec

Y₁₆ = position of rocket at t = 16 sec

a' = acceleration of rocket

t = time of travel = 16 sec

v₁₆ = velocity of rocket at t = 16 sec

using the equation

v₁₆ = v₀ + a' t

v₁₆ = 0 + 16 a'

v₁₆ = 16 a'

position is given as

Y₁₆ = Y₀ + v₀ t + (0.5) a' t²

Y₁₆ = 0 + 0 (16) + (0.5) a' (16)²

Y₁₆ = 128 a'

consider the motion of rocket from t = 16 sec   to t = 20 sec

v₁₆ = velocity of rocket at t = 16 sec  =  16 a'

Y₁₆ = position of rocket at t = 16 sec  = 128 a'

Y₂₀ = position of the rocket a t = 20 sec , = 5100 m

a = acceleration = - 9.8 m/s²

Δt = time interval = 4 sec

Using the equation

Y₂₀ = Y₁₆ + v₁₆ Δt + (0.5) a Δt²

5100 = 128 a' + (16 a') (4) + (0.5) (- 9.8) (4)²

a' = 27 m/s²

b)

v₂₀ = velocity of rocket at t = 20 sec  =  ?

Using the kinematics equation

v₂₀ = v₁₆ + a  Δt

v₂₀ = 16 a' + (- 9.8) (4)

v₂₀ = 16 (27) + (- 9.8) (4)

v₂₀ = 392.8 m/s