# Test problem

1. Feb 18, 2005

we had the following test question last week in my physics class:

A rocket is launched at an angle of 7.5 degrees and reaches a maximum height of H. To make the rocket reach 2*H what needs to be done.

a) double the launch angle and initial velocity

b) half the range

there were two other options but i can't remember them.

The only hint he would give us (after the test of course ) was that we had to find a general solution. ONLY ONE PERSON GOT THE ANSWER RIGHT AND HE GUESSED IT.the test was multiple choice of course.

Any ideas ?

2. Feb 18, 2005

### learningphysics

Derive the formula for maximum height H in terms of v (initial velocity), theta (launch angle) and g.

Was there a sqrt(2) in any of the answers?

Last edited: Feb 18, 2005
3. Feb 18, 2005

### Staff: Mentor

Start by finding the maximum height H as a function of launch angle and initial speed.

4. Feb 18, 2005

no there was not a sqrt(2) as an answer choice. i don't know how to derive that equation but i'll give it a shot

5. Feb 18, 2005

### vincentchan

use conservation of energy... the kinetic energy is 1/2mv^2 and the potential energy is mgH, if you double the height(H), what will the initial velocity(v) be?

6. Feb 18, 2005

### Andrew Mason

This is a rocket, not a projectile, but I don't think it really makes any difference. What you want to do is give the rocket maximum height for the amount of fuel it has. In order to do this you put all the energy into gaining height so that it has 0 kinetic energy at maximum height. The only way to do this is to shoot it straight up.

AM

7. Feb 18, 2005

### learningphysics

I meant the maximum height the projectile reaches when it's launched with a particular velocity and angle.... the maximum y coordinate of the trajectory.
The question mentions that the maximum height when the angle is 7.5 is H. (so we're not talking about the maximum possible height at any angle, but the maximum y-coordinate at a particular angle and velocity).

Also, I just wanted to mention that the way I understand the question... it's asking for what needs to be changed to make the "maximum height" during the trajectory 2H. If we double the angle and the velocity, it will definitely reach a height of 2H, but the maximum height will be more than that.

EDIT: Sorry. I just understood what Andrew meant in his post. I grossly misunderstood the problem. Apologies.

Last edited: Feb 18, 2005
8. Feb 18, 2005

thank you i see now

9. Feb 18, 2005

### Staff: Mentor

Good point. I just assumed that the author of the question was sloppy in his language and intended this to be a projectile motion problem, not a rocket problem. I interpreted the problem the same way that learningphysics did.

10. Feb 18, 2005

this is supposed to be projectile motion. imagine a water rocket.

11. Feb 18, 2005

### Andrew Mason

I am not sure there is a significant difference between a projectile and a rocket, apart from the shape of the trajectory. When the rocket fuel runs out, the rocket becomes a projectile.

I am not sure what is meant by a 'general solution' to a specific question. What is required to enable the rocket to reach a height of 2H is to increase its vertical kinetic energy by a factor of at least 2 ($sin\theta = \sqrt{2}sin(7.5)$). But since a 'general solution' was sought, it seems to me that the general way to make it 'reach' a greater height is to shoot it straight up. I guess the question is: what were the other possible answers.

AM