Dart Fired from a Dart Gun

  • Thread starter Becca93
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  • #1
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Homework Statement


I need help with the very last of a series of questions:


The potential energy stored in the compressed spring of a dart gun, with a spring constant of 66.00 N/m, is 0.76 J. Find by how much is the spring is compressed.
My correct answer: 0.152 m

A 0.190 kg dart is fired straight up. Find the vertical distance the dart travels from its position when the spring is compressed to its highest position.
MCA: 0.408 m

The same dart is now fired horizontally from a height of 4.10 m. The dart remains in contact until the spring reaches its equilibrium position. Find the horizontal velocity of the dart at that time.
MCA: 2.83 m/s

Find the horizontal distance from the equilibrium position at which the dart hits the ground.



The Attempt at a Solution



I have no idea how to solve this last question. I thought for a moment that I would use potential energies to find something I could plug into a kinematics equation, but the height up and down would be the same, negating mgh, there is no spring involved at this point, negating (1/2)kx^2, and the mass and horizontal velocity do not change.

Any suggestions on how to proceed?
 

Answers and Replies

  • #2
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Hey Becca93.
I think this is a projectile motion problem. Horizontal Velocity is the horizonal component i.e. Vcosθ. Horizontal acceleration is 0m/s2 as velocity is constant throughout the flight, and time of flight is tseconds.

From then on it's as easy as using the equations of constant acceleration to find the horizontal distance.

(Hint: V0 = Vcosθm/s, and Vf = 0m/s.)
 
Last edited:

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