Help with these rocket launch questions please

[takelight2]

Homework Statement

A toy rocket moving vertically upward passes by a 2.2 m -high window whose sill is 9.0 m above the ground. The rocket takes 0.17 s to travel the 2.2 m height of the window.

1. What was the launch speed of the rocket? Assume the propellant is burned very quickly at blastoff.

2. How high will the rocket go?

Relevant Equations

SUVAT equations.

I know that a=-9.8, I am having trouble aplying the motion equations.

For example, I can't use equations that have velocity (either initial or final), so I can rule out those equations. I am then left with no equations to use. I am extremely stumped.

 

[Master1022]

The rocket takes 0.17 s to travel the 2.2 m height of the window.

Is this saying that it took 0.17s to get to the window? Or is it saying that the window has a height of 2.2 m (i.e. bottom is 9 m off the ground and top is 11.2 m above the ground) and it took 0.17 s to get from bottom of window to top of window?

I only ask as I have seen problems that use the latter before

[EDIT]: I have re-read the problem and it looks like it is the latter?

 

[Master1022]

Homework Statement:: A toy rocket moving vertically upward passes by a 2.2 m -high window whose sill is 9.0 m above the ground. The rocket takes 0.17 s to travel the 2.2 m height of the window.
1. What was the launch speed of the rocket? Assume the propellant is burned very quickly at blastoff.

I know that a=-9.8, I am having trouble aplying the motion equations.

Okay good, so you know $a$ and we also have $u$, $v$, $t$ and $s$ which are unknown. For the first part, you have too many unknowns to get any information from a single equation, but what if you set up a system of equations with multiple unknowns that could be used to solve for some of those points.

Spoiler: Hint 1

Can we set up 2 equations to solve for the time at which the window is reached $t$ and the initial speed $u$?

Spoiler: Hint 2

Can use $s = ut + \frac{1}{2}at^2$ at the bottom and the top of the window to form those system of equations? We know $s$ at the top and the bottom. We know $a$.

Spoiler: Hint 3

We know that the time $t_{top} = t_{bottom} + 0.17$

Hope that is of some help

 

[takelight2]

Is this saying that it took 0.17s to get to the window? Or is it saying that the window has a height of 2.2 m (i.e. bottom is 9 m off the ground and top is 11.2 m above the ground) and it took 0.17 s to get from bottom of window to top of window?

I only ask as I have seen problems that use the latter before

[EDIT]: I have re-read the problem and it looks like it is the latter?

Yup its the latter.

 

[Master1022]

Yup its the latter.

Thanks for the clarification. Then you can set up a system of equations (as suggested in the above post).

Hope that provides a starting point. If not, will be happy to clarify.

 

[takelight2]

Okay good, so you know $a$ and we also have $u$, $v$, $t$ and $s$ which are unknown. For the first part, you have too many unknowns to get any information from a single equation, but what if you set up a system of equations with multiple unknowns that could be used to solve for some of those points.

Spoiler: Hint 1

Can we set up 2 equations to solve for the time at which the window is reached $t$ and the initial speed $u$?

Spoiler: Hint 2

Can use $s = ut + \frac{1}{2}at^2$ at the bottom and the top of the window to form those system of equations? We know $s$ at the top and the bottom. We know $a$.

Spoiler: Hint 3

We know that the time $t_{top} = t_{bottom} + 0.17$

Hope that is of some help

Thanks for the help. I tried the things you said, and ended with the following system of equations:

$9 = ut + \frac{1}{2}(-9.8)t^2$

$11.2 = u(t+0.17) + \frac{1}{2}(-9.8)(t+0.17)^2$

Now, I don't really find a way to solve this system, and it seems to be overly complicated? Am I doing something wrong??

 

[gmax137]

You are on the right track. Some messy algebra will let you solve for initial velocity u and time t. I kept it symbolic until the end (I used g rather than 9.81 and δ for 0.17) .

 

[takelight2]

You are on the right track. Some messy algebra will let you solve for initial velocity u and time t. I kept it symbolic until the end (I used g rather than 9.81 and δ for 0.17) .

It would be -9.8 right? since its opposite the direction of motion..?

 

[Master1022]

It would be -9.8 right? since its opposite the direction of motion..?

Yes - I think @gmax137 was just referring to magnitude.

$9 = ut + \frac{1}{2}(-9.8)t^2$

$11.2 = u(t+0.17) + \frac{1}{2}(-9.8)(t+0.17)^2$

Now, I don't really find a way to solve this system, and it seems to be overly complicated? Am I doing something wrong??

As mentioned above, this looks correct (and working algebraically might help to prevent any numerical errors along the way). Now you have a two equations with two unknowns. Can you find a way to eliminate a variable from either equation?

 

[gmax137]

would be -9.8 right? since its opposite the direction of motion..?

Yes - I think @gmax137 was just referring to magnitude.

Yes - Sorry for the confusion.

Were you able to solve the two equations to find the initial velocity?

This approach hinges on the meaning of this:

Assume the propellant is burned very quickly at blastoff.

Can you see why? What simplifying assumptions have we made in setting up the two equations, based on this?

 

[jbriggs444]

There is an easier approach which does not involve simultaneous equations.

We know a distance covered and an elapsed time. That means that we can compute an average velocity.

We know that acceleration is constant. That means that the average velocity is also the velocity halfway through the interval. That is enough information to determine the starting and ending velocities from there.

 

[gmax137]

There is an easier approach which does not involve simultaneous equations.

We know a distance covered and an elapsed time. That means that we can compute an average velocity.

We know that acceleration is constant. That means that the average velocity is also the velocity halfway through the interval. That is enough information to determine the starting and ending velocities from there.

I will have to try working it through that way. I'm not so sure.