How High Does a Rocket Go in a Physics Experiment?

[Rockdog]

In an experiment in physics class, a model solid-fueled rocket is fired vertically with an acceleration of 5 m/sec2 for 7 seconds. After that time, it's fuel is exhausted and it continues upwards as a free fall particle. (Take upwards to be the positive direction.)

a) What is the maximum altitude the rocket reaches?

b) What is the rocket's acceleration 8.5 seconds after launch?

c) What is the total time of flight?

d) What is the rocket's total displacement during the entire flight?

I'm only stuck on letter a right now, but if I need further help, I'll let you guys know.

For part a, what I tried doing is using this equation,
x-x0=vo(t)+(1/2)a(t)^2
where x0=0
vo=0
a=5 m/sec^2
t=7

I get x= 122.5 meters. Computer won't take it, so I'm scratching my head to no end. Any ideas to help me get started?

 

[krab]

There are 2 stages: (1) it accelerates upward (a=5m/s/s), (2) it is in free fall (a=-9.8m/s/s). You only dealt with the first stage.

 

[HallsofIvy]

RockDog- you made this same mistake in your previous post!

In the previous post, you calculated distance as if the car were decelerating the entire time when the problem specifically said that it only accelerated for 5 seconds and then continued at constant speed.

Here the problem specifically says " it continues upwards as a free fall particle." but you only calculated its height when the rocket engine cut off!

I don't know if we can allow much more of that!

 

[Rockdog]

Ok, I admit it. When it comes to physics, I am poor at it, but you guys are helping me become better. Isn't that scary? :-)

Anyhow, I figured out 3/4 of the problem so far, and I'm stuck on part c.

I need to figure out the total time of the whole flight.
It takes 7 seconds for the rocket to go from ground to the point it cuts off it fuel.

Then I figured out how much time it took for the rocket to go from the point of shutting its fuel off to the maximum height of the parabola.

I also figured out that the distance from ground to engine shutting off is 122.5 m.

I figured out initial velocity is going to be v=v0+at
v=0, a=-9.8m/s/s t=7 sec
So v0= 35 m/s.

Now I'm having trouble finding the time it takes to go back to the ground.

I tried using y-y0=v0(t)+(1/2)*a*(t)^2
where
y=0
y0=122.5m
v0=35m/s
a=-9.8 m/s/s
and I get a t= 9.72sec or -2.57 sec.

So total time would be 7sec +9.72sec=16.72 secs, but the computer won't take it. Any ideas?

 

[HallsofIvy]

" also figured out that the distance from ground to engine shutting off is 122.5 m."

Yes, that's correct.

"I figured out initial velocity is going to be v=v0+at
v=0, a=-9.8m/s/s t=7 sec

So v0= 35 m/s."

I don't see how you get that. Assuming you put v= 0, a= -9.8, t= 7 into v= v0+ at, you get 0= v0- 9.8(7) so v0= 9.8(7)= 68.6, not 35.
I think you actually did it correctly but wrote it down wrong. While the rocket engine is firing a= 5. v0= 0 so v= 0+ 5(7)= 35 and you can use THAT as v0 after the engine shuts down.

AFTER the rocket engine has shut off, the only acceleration is that of gravity: -9.8 so v= 35- 9.8t (t is now measured from engine cutoff, not initial launch). From that we get h= 35t- 4.9t²+ 122.5 (since the height was 122.5 m at engine cutoff, t=0).

The maximum height occurs when v= 0: 35- 9.8t= 0 gives t= 35/9.8= 3.57 seconds. That height will be h= 35(3.57)- 4.9(3.57)²+ 122.5= 185 meters.

Finally, the flight will be over when the rocket hits the ground:
h=0. Solve 35t- 4.9t²+ 122.5= 0 to find the time AFTER CUTOFF that that occurs. You will need to add the inital 7 seconds to find the length of the entire flight.