# Kinematics toy rocket problem

1. Aug 30, 2010

### zcd

1. The problem statement, all variables and given/known data
A toy rocket moving vertically upward passes by a 2.2 m high window whose sill is 9.0m above the ground. The rocket takes 0.17s to travel the 2.2m height of the window. What was the launch speed of the rocket, and how high will it go? Assume the propellant is burned very quickly at blastoff.

2. Relevant equations
$$y=-4.9t^2+v_0t+y_0$$

3. The attempt at a solution
I start by assuming the rocket launched from the ground, so the initial height is 0. The height at an unknown time y(t0)=9.0=-4.9t02+v0t0, and the height .17s after is y(t0+.17)=2.2=-4.9(t0+.17)2+v0(t0+.17). If I take the difference, I end up with:
2.34161=-1.666t0+0.17v0.

If my reasoning wasn't wrong here, it seems I'm missing a piece of information to finish the problem. How would I find the exact point in time t0 where the rocket just reaches the window sill?

2. Aug 30, 2010

### rl.bhat

y(t0+.17)=2.2=-4.9(t0+.17)2+v0(t0+.17).

The above equation should be

y(t0+.17)=2.2 + 9 =-4.9(t0+.17)^2+v0(t0+.17).

Now solve the equations to find vo and to.

3. Aug 30, 2010

### zcd

I don't see how I can solve the equations. I don't know the value of t0

4. Aug 30, 2010

### rl.bhat

9.0=-4.9t0^2+v0t0,.....(1)

2.2 + 9 =-4.9(t0+.17)^2+v0(t0+.17).....(2)

(2) - (1)

2.2 = -(4.9)[(to + .17)^2 - to^2] + vo*0.17

2.2 = -(4.9)[2*o.17*to + (0.17)^2] + vo*0.17

0.17*v0 = 2.2 + (4.9)[2*o.17*to + (0.17)^2]

Find vo and substitute is eq(1) and solve for to.

5. Aug 30, 2010

### zcd

I didn't think of that. Thanks for the help!

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