Test rocket acceleration physics problem

[elizabethg]

I need help, i tried this problem many times and i keep getting the wrong answer.
A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of 79.2 m/s at ground level. Its engines then fire and it accelerates upward at 4.10 m/s2 until it reaches an altitude of 950 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of -9.80 m/s2. (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?
___________s
(b) What is its maximum altitude?
_________km
(c) What is its velocity just before it collides with the Earth?
_________________m/s

I tried finding the time with the equations x=.5at^2+Vot+Xo. I get nowhere, please help. I plug in numbers and still get the wrong answer.

I tried doing different equations for going up, the time it stops in the air, and when it comes down, i just don't know which equation is right, though i do know you need three separate times to find the total time.

 

[jamesrc]

Well, that is the right equation to use here (I like using y for vertical motion rather than x, but that's not important). Let's look at each part (using only the kinematic equations for constant acceleration motion you know):

Part 1: (engines on)

Your equation is: x₁ = x_o + v_ot₁ + .5at₁²

What do you know? Make sure you understand the meaning of each variable rather than just trying to plug things in.

x_o = 0 (starts at ground level)
v_o = 79.2 m/s (its "initial" speed, i.e. the speed it has (due to the catapult) at time t = 0)
x₁ = 950 m (the altitude it reaches when the engines shut down)
a = 4.1 m/s/s (the acceleration due to the thrust provided by the rocket)

Now solve that equation for t₁. You'll also want to know the velocity of the rocket at t₁. That is easily found:

v₁ = v_o + a*t₁

Part 2: (simple ballistic motion upward to its maximum height)

The same equation will work for us here (subscripts changed to reflect that this is the second part of our problem):
x₂ = x₁ + v₁t₂ + .5at₂²

but we've got two unknowns in there. Here's a quicker way to get to the answer:

v_2^2 = v_1^2 + 2a\Delta x

Remember that in this part of the problem, a = -9.8 m/s/s
v₂ is 0, since this is at the peak of the motion

Solve for Δx, noting that \Delta x = x_2 - x_1, which allows you to solve for the maximum altitude acheived = x₂

Since we also want to know the time here, calculate it using:

v₂ = v₁ + at₂

Note that as I've defined things here, t₂ is the elapsed time during part 2 ot the trip

Part 3: (falling to the ground)

Last leg of the trip. Back to our first equation (which seems to be the most useful right now):

x₃ = x₂ + v₂t₃ + .5at₃²

and solve for t₃

a is still -9.8m/s/s and x₃ is 0 (since we land on the ground)

To find the total time in the air, just add of the 3 individual times.

To find the velocity just before it hits the ground, go back to the equation:

v₃ = v₂ + at₃

That's a long problem and I hope I haven't made any mistakes. If you follow the logic, you should be able to decide for yourself if it's right or not. I hope that helps.

 

[Fredrik]

Of course someone else posted an answer while I was writing mine. OK, this may be useless now, but I might as well post it since I've already written it. Here it is:

If you solve for t and plug in the numbers, you get the time at which the engine failed. Now you have to figure out the velocity it had at that time. This you do with another equation (the derivative of the one you showed). When you've done that, use the first equation again to find the time it takes the rocket to fall to the ground. The sum of the two times will be the answer to a.

You will have to use both equations again to find the answer to b. And if you can handle that, I'm sure you'll have no problems with c.

 

[Beer-monster]

Do you know what the numerical should be?

I tried it myself and would like to know if I'm right ^__^