# Crazy Projectile Motion Problem Coming Right At Yah!

1. Oct 7, 2011

### loopsnhoops

1. The problem statement, all variables and given/known data

Find the initial velocity and the maximum height of the rocket.

Angle = 90˚
Horizontal distance = 0 m
Time = 6.55 seconds

2. Relevant equations

Suvat equations.

3. The attempt at a solution

s = ut + 1/2a(t^2)
s = 6.55u
0/6.55 = u

???

2. Oct 7, 2011

### Quantum Mind

It would help to write down the known quantities and the unknowns:

v, t and a are known. Since the angle is 90°, there is no horizontal component involved. The unknowns are u and S. Thinking of equations where only one of the unknowns is used will help.

3. Oct 7, 2011

### loopsnhoops

v = u + at?

but how does that work I would get

0 = 0 + 0(6.55)

4. Oct 7, 2011

### Quantum Mind

If both initial and final velocities are zero and the acceleration is zero, how does the rocket move?

u cannot be zero since that is what you are supposed to find . What is the velocity (final velocity) of the rocket at its maximum height? What is its acceleration?

5. Oct 7, 2011

### loopsnhoops

Thanks so far i got u which is 32.75 I think but I don't know how to find the maximum height.

6. Oct 7, 2011

### Cipherflak

Funny that they call it a rocket. Rockets are usually self propulsive by hurling mass backwards. This is a normal projectile, right? :)

7. Oct 7, 2011

### Cipherflak

s = vt + 1/2a(t^2).

here you have a solution for s. if you plug in s=0 (the ground where the projectile starts), you probably have two different t that satisfy this. One is obviously t=0, but the other?.

As for the maximum height; well, what you have here is distance as a function of time s(t). What about finding the maximum value of this graph? ;)