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jwxie
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rocket problem (time is wrong!)
A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of 80.0 m/s at ground level. Its engines then fire and it accelerates upward at 4.00 m/s^2 until it reaches an altitude of 1000m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of -9.80 m/s^2. (You will need to consider the motion while the engine is operating separate from the free-fall motion.)
a) How long is the rocket in motion above the ground?
b) What is its maximum altitude?
c) What is its velocity just before it collides with the Earth?
assuming constant a
Vf = Vi + at
Xf = Xi + 1/2(Vi + Vf)t
Xf = Xi + Vit + 1/2 at^2
V = Vi + Vf / 2
I understand the problem very well, but my answer didn't come out right. Let's do question A.
We find the final velocity at the moment the engine right before or at that instance it fails. So Vf (eng) = Vi (free fall upward).
Now we are given 1000m for the engine, so we can solve it this way
Vf^2 = Vi^2 + 2a(Xf - Xi) and let Xf = 1000m, and a = +4m/s^2 , and I got
Vf = 120m/s
Now solve for t (how long it takes to get up to 1000)
Vf = Vi + at
I got 120 = 80 + 4t, so t = 10s
So we have Vi (free fall upward) = 120m/s
let's solve for t
Vf = Vi + at
0 = 120 + -9.8 * t
t = 12.2
According to the book, the answer is 41.0s but now i only have 22.2 second
How come?
Homework Statement
A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of 80.0 m/s at ground level. Its engines then fire and it accelerates upward at 4.00 m/s^2 until it reaches an altitude of 1000m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of -9.80 m/s^2. (You will need to consider the motion while the engine is operating separate from the free-fall motion.)
a) How long is the rocket in motion above the ground?
b) What is its maximum altitude?
c) What is its velocity just before it collides with the Earth?
Homework Equations
assuming constant a
Vf = Vi + at
Xf = Xi + 1/2(Vi + Vf)t
Xf = Xi + Vit + 1/2 at^2
V = Vi + Vf / 2
The Attempt at a Solution
I understand the problem very well, but my answer didn't come out right. Let's do question A.
We find the final velocity at the moment the engine right before or at that instance it fails. So Vf (eng) = Vi (free fall upward).
Now we are given 1000m for the engine, so we can solve it this way
Vf^2 = Vi^2 + 2a(Xf - Xi) and let Xf = 1000m, and a = +4m/s^2 , and I got
Vf = 120m/s
Now solve for t (how long it takes to get up to 1000)
Vf = Vi + at
I got 120 = 80 + 4t, so t = 10s
So we have Vi (free fall upward) = 120m/s
let's solve for t
Vf = Vi + at
0 = 120 + -9.8 * t
t = 12.2
According to the book, the answer is 41.0s but now i only have 22.2 second
How come?
