The potential energy stored in the compressed spring of a dart gun, with a sprin

Question

The potential energy stored in the compressed spring of a dart gun, with a spring constant of 45.00 N/m, is 1.320 J. The spring is compressed by 0.2422 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. Use your equations of motion under constant acceleration (in this case, the surface gravity of Earth).

The same dart is now fired horizontally from a height of 1.10 m. The dart remains in contact until the spring reaches its equilibrium position. Find the horizontal velocity of the dart at that time.

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

Explanation / Answer

Energy transmitted to the dart = 1.32J
mgh = 1.32
Hence h = 1.32/(0.19*9.8) = 0.709m

Now the dart is fired horizontally
The vertical component of velocity initially is 0
Hence the time taken for the dart to fall = (2h/g) = 0.4738s

Initially velocity of dart be v

0.5mv^2 = 1.32

v = 3.728m/s

Hence the horizontal distance travelled = 3.728*0.4738 = 1.766m

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