Spring constant, find height of projectile

  • Thread starter subopolois
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  • #1
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Homework Statement


A 48 gram dart is shot vertically upwards from a catapult with a spring constant of 350 N/m. The catapult is initially stretched from the equilibrium point by 21 cm. What is the height above the starting point reached by the dart?


Homework Equations


KE=PE, rearranging to find h gets h= v^2/2g

find the velocity as it leaves the catapult as v= d(sqrt(k/m))

The Attempt at a Solution


v= 0.21m (sqrt(350Nm/0.048kg))
= 17.93 m/s

h= (17.93m/s)^2 / 2(9.81m/s^2)
= 16.38 m

does this seem right?
 

Answers and Replies

  • #2
TSny
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Your work looks good. But did you answer the specific question that was asked?
 
  • #3
TSny
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Sorry, subopolois. I read your post too quickly.

When you calculated the speed at which the dart leaves the catapult, did you take into account the increase in gravitational potential energy of the dart? How did you get the formula v = d√(k/m)?
 
  • #4
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I did not, how would i do that?
I did take gravity into account when i found the height...
 
  • #5
TSny
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You could find the speed at which the dart leaves the catapult by using conservation of energy. You will need to include kinetic energy, gravitational potential energy, and elastic potential energy.

There is a better approach to the problem. You don't need to find the speed at which the dart leaves the catapult. Is energy conserved overall between the point of release and the point where the dart reaches the highest point?
 

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