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Spring gun problem without a spring constant. 
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#1
Nov506, 04:20 PM

P: 30

Hello:
I'm stuck on this following problem. A spring gun, held horizontally 1.6 meters above the ground, fires a 0.07 kilogram ball so that it lands a horizontal distance of 2.3 meters away. If the gun is pointed straight up, and the same ball is fired, how high (in meters) will it rise? I know that: (Potential gravity) = (mass) x (gravity) x (height) Also: (Spring potential) = (1/2) x (k) x (x^2) I'm not really sure how to tie the two together... Also, I'm thinking of using the conservation of energy formula: (final kinetic + final potential gravity + final spring potential) = (initial kinetic + initial potential gravity + initial spring potential) but I don't know how to get to the spring constant (k) with the given information in this problem. Any help would be appreciated. 


#2
Nov506, 06:00 PM

P: 312

The spring has nothing to do with it, youre better off with kinematics. Figure out the initial velocity in the x for the first fire



#3
Nov506, 08:13 PM

P: 30

I considered the kinematics of this problem. I used two kinematics equations, one solving for time (using the height of the gun off the ground) and the other considering gravity and set them equal to each other. That's how I got velocity (approx 4 m/s). Then I applied the conservation of energy theorem for the moment right after the bullet leaves the gun (initial) right before it hits the ground (final). This is where the spring constant and the x (spring compression) come into play. Are you sure this problem can be done by completely disregarding the spring? Even though I got the velocity, I still don't know how to deal with not having the spring compression.



#4
Nov506, 10:40 PM

P: 126

Spring gun problem without a spring constant.
TurdFerguson is right, this problem would be the same if you were just throwing the object, so pretend there is no spring at all.
I'm a little rusty, but isn't the pot. energy mgh? 


#5
Nov506, 10:54 PM

P: 30

(gravitational potential energy) = m x g x h I'm still stuck on this one. I'm not sure what to do after finding the velocity. 


#6
Nov1006, 12:20 AM

P: 30

This problem was solved by finding time with a kinematic equation on the yaxis. Once that was found, I put the time into an another kinematic equation in order to find velocity on the xaxis. This velocity was used with a conservation of energy equation. So, through this final conservation equation the height was found.
Thank you for the hints! 


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