# Spring launched

1. Jul 17, 2013

### Zondrina

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

A spring is launched from the ground at 2.3 m/s t an angle of 78° to the ground. What is the max height of the spring? How far did the spring travel?

2. Relevant equations

$\vec{v}_R = 34 m/s$
$\theta = 78°$

3. The attempt at a solution

First things first, get my components :

$\vec{v_v} = 34sin(78°) = 33.3 m/s [Up]$ [ Vertical velocity ].
$\vec{v_H} = 34cos(78°) = 7.1 m/s [Forward]$

Hmm looks like I need time. So lets use $\vec{Δd}_V = \vec{v_v}Δt - (4.9)(Δt)^2$. We know that the overall vertical displacement will be 0 because the spring returns to rest due to gravity. So simplifying :

$0 = (33.3 - 4.9Δt)Δt$

So either $Δt = 0$ which we know we can throw out, or $Δt = \frac{-33.3}{-4.9} = 6.8 s$.

Now we know how long the spring was in the air. So because of acceleration due to gravity, we know that the max height of the spring will be achieved at $Δt_S = Δt/2 = 3.4 s$.

To find the actual maximum height we calculate :

$\vec{Δd}_V = \vec{v}_v Δt_S - (4.9)(Δt_S)^2 = (33.3)(3.4) - (4.9)(3.4)^2 = 56.58 m/s [Up]$

To find out how far the spring traveled in total, another small calculation :

$\vec{Δd}_H = \vec{v}_H Δt = (7.1)(6.8) = 48.3 m [Forward]$.

I think that should be it. Loving these word problems.

EDIT : Just noticed this should be in intro phys, I must have clicked adv by accident. If a mod could move it there it would be appreciated.

2. Jul 17, 2013

### Dick

I'm not sure why your statement says velocity 2.3m/s and you finally used 34m/s. But other than that it looks pretty ok.

3. Jul 17, 2013

### Zondrina

Oh wow haha. Been awake since 5am, taking its toll i guess, but these word problems are too fun to solve.

Quantifying the physical.

Okay though, as long as the process is fine, that's all I really care for.