## Gravitional Potential Energy and Elastic Potentional Energy

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

1) A rifle shoots a spring of mass 0.008kg and with a spring constant of 350 N/m. You wish to hit a target horizontally a distance of 15 m away by pointing the rifle at 45 degrees above the horizontal. How far should you extend the spring in order to reach the target?

2. Relevant equations
1) d=v*t, v2=v1+a*t, d=v1(t)+(0.5)(a)(t)

3. The attempt at a solution
im lost :(

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

A bungee cord need to transfer 2,000,000 J of energy. A 10 kg mass extends the bungee cord 1.3m. What is the maximum extension of the bungee cord?

2. Relevant equations
F=kx, E=(0.5)(k)(x^2) W=F*d

3. The attempt at a solution
lost again :(

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 you need to show what you've tried in the question

 Quote by cupid.callin you need to show what you've tried in the question
sadly, i dont even know where to start.

## Gravitional Potential Energy and Elastic Potentional Energy

Try to start by finding the maximum height that the spring must reach. Then try to consider changes in energy.

 Quote by AlexChandler Try to start by finding the maximum height that the spring must reach. Then try to consider changes in energy.
can you start me off with few steps please?

 Quote by y201 can you start me off with few steps please?
Sure. Try to combine the kinematics equations by eliminating the time in order to reach the trajectory equation. It should look like this.

$$y = tan \theta_0 x - \frac{g cos^2 \theta_0}{2 V_0^2} x^2$$

Then you should be able to find the maximum height.

 Quote by AlexChandler Sure. Try to combine the kinematics equations by eliminating the time in order to reach the trajectory equation. It should look like this. $$y = tan \theta_0 x - \frac{g cos^2 \theta_0}{2 V_0^2} x^2$$ Then you should be able to find the maximum height.
this equation is for the second question right?

 Quote by y201 this equation is for the second question right?
No it is for the spring problem.

 you can also do it the traditional way find the time of flight using y = ut + .5gt2 now as horizontal speed donot change throughout the motion: distance you need to reach, d = uX * t so you have uX viz ucos45 now use energy conservation .5 kx2 = .5mv

 Tags energy, force, maximum, spring, transfer