Answer: Calculating Velocity of 1000kg Rocket After Launch

[johnlu66]

Homework Statement

A 1000kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16s, then the motor stops. The rocket altitude 20s after launch is 5100m. You can ignore any effects of air resistance.

What was the rocket's acceleration during the first 16s?

What is the rocket's speed as it passes through a cloud 5100m above the ground?

i wasnt able to get started it...need solutions asap.

2. Homework Equations

v=vo+at
x=volt+0.5at^2
v^2=vo^2+2ad

3. The Attempt at a Solution
i don't know if vo is 0 or not, but i got 34 but its aint right.

 

[LowlyPion]

Homework Statement

A 1000kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16s, then the motor stops. The rocket altitude 20s after launch is 5100m. You can ignore any effects of air resistance.

What was the rocket's acceleration during the first 16s?

What is the rocket's speed as it passes through a cloud 5100m above the ground?

i wasnt able to get started it...need solutions asap.

2. Homework Equations

v=vo+at
x=volt+0.5at^2
v^2=vo^2+2ad

3. The Attempt at a Solution
i don't know if vo is 0 or not, but i got 34 but its aint right.

Welcome to PF.

V_o is zero when it is fired.

Since after 16 seconds the rocket stops, you have 2 phases to the problem. Uniform acceleration for 16 seconds and then gravitational trajectory thereafter.

 

[johnlu66]

Welcome to PF.

V_o is zero when it is fired.

Since after 16 seconds the rocket stops, you have 2 phases to the problem. Uniform acceleration for 16 seconds and then gravitational trajectory thereafter.

well, how about final velocity? is it also zero? it seems not but i thought it should be zero after the 20 sec...omg

ok here is what i have changed,
for t = 16
v = a(16)
d= 0.5*a*(16)^2

for t = 4
v = 16a(4)
5100-d = 16a(4) - 9.8 (0.5)(4^2)

5100 = 128a + 64a - 78.4

solve for a? i got a = 26.97 this time