# Rocket travels along parabolic path (RECTANGULAR COMPONENTS)

1. Feb 3, 2010

### patflo

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

When a rocket reaches an altitude of 40m it begins to travel along the parabolic path (y-40)^2=160x, where the coordinates are measured in meters. If the component of velocity in the vertical direction is constant at vy=180m\s, determine the magnitudes of the rockets velocity and acceleration when it reaches an altitude of 80m.

2. Relevant equations

position: r=x*i+y*j (where i and j are the unitvector)
velocity: v=dr\dt
acc: a=dv\dt

3. The attempt at a solution

lets say i tried for about 4 hours to get a concept to solve this. :(

first i put in in the equation for the motion path for y=80 to get x=10

second: position vector for point A (x=0,y=40) and point B (x=10 y=80)

third: to get the distance betwee A and B i subtract the 2 position vectors.

fourth: v=dr\dt to get the velocity

mazbe its the wrong way because the answer should be v=201 m\s and mines was 254 m\s.

i dont know where my mistake is...is the way of thinking right?

2. Feb 3, 2010

### aim1732

I don't how you found out dr/dt without separating out the equations for motion on x,y-axes w.r.t time from the trajectory equation.
Velocity on y axis is constant so y=180t.
Put in trajectory eqn. to eliminate y then differentiate w.r.t. x. This will be velocity on x-axis. Add vectorially for complete velocity.
Or you could directly differentiate the trajectory equation w.r.t time. dy/dt is known, dx/dt can be found out and then added vectorially.

3. Feb 3, 2010

### Gokul43201

Staff Emeritus
Notice that eqn#2 has a variable in it that is defined in eqn#1. If you plug this definition from #1 into #2, what equation do you get for the velocity?

4. Feb 3, 2010

### patflo

ok i tried out this:

d/dt (y-40)²=d/dt 160x

=> dx/dt=405t-90

v=vx*i+vy*j

v=(405t-90)*i+180j

but how can i find the magnitude? since in the equation for v there is still the variable "t".

5. Feb 3, 2010

### Gokul43201

Staff Emeritus
Can you now find out how much time it takes for the projectile to reach a height of 80m?

Last edited: Feb 3, 2010
6. Feb 3, 2010

### patflo

for the equation of path i put in y=180t and x=10 (when y=80 x=10)

then i get a quadratic function in t.

the result for t= 0.4s

i put that in v and get v=72i+180j

for magnitude: v=193,7 m/s

but as result there should be 201 m/s

7. Feb 4, 2010

### aim1732

t=(distance on y-axis)/(Speed on y-axis)
This is 40/180=0.22s.

8. Feb 4, 2010

### patflo

thank you very much!!