# Homework Help: Rocket with only time

1. May 26, 2005

### jenduncan

A 1000 kg rocket is launched straight up. The rocket motor overcomes gravity to provide the rocket with a net constant acceleration upward for 16 seconds. The the motor shuts off. 4 seconds later the rocket's altitude is 5100 meters.
What is the rockets acceleration during the 1st 16 seconds?

0=v+-9.8 m/s^2 *4s = 39.2m/s but I'm off by a factor of 10 for the velocity at 5100 m

Last edited: May 26, 2005
2. May 26, 2005

### inha

What have you done so far?

3. May 26, 2005

### inha

What exactly are you going for here? The assigment does not say that the rocket is at halt at 5100m.

4. May 26, 2005

### HallsofIvy

Call the acceleration asked for "a". Then, for the first 16 seconds, v(t)= at and
h(t)= (a/2)t2 (v(t) is the velocity t seconds after lift off, h(t) is the height t seconds after lift off). From that you can find the velocity, v(16), and height h(16) when the rocket shuts off as a function of a.

Now the acceleration is just -g: v(t)= v16- gt and h(t)= h16+ v16- (g/2)t2, where v16 and h16 are the speed and height of the rocket when the rocket shut off. t is, of course, the time after the rocket shut off so set h(4)= 5100 and you will have an equation in a. Solve that equation.

Notice that the mass of the rocket is irrelelvant.

Last edited by a moderator: May 26, 2005