You are trying to design a spring gun that is able to launch a 7.0 g marble to a

Question

You are trying to design a spring gun that is able to launch a 7.0 g marble to a vertical height of 22 m (measured from the starting position of the marble). The specifications for the gun state that when it is loaded, the spring will be compressed 8.0 cm from its relaxed length. (When the gun is fired, the ball is released when the spring reaches its relaxed length.) What should the spring's spring constant be?

V(r) = (1/2)k(r-r0)^2
where k=spring constant and r0=initial distance of separation

Explanation / Answer

22*2*9.81 = v^2 =>v = 20.77594763m/sec 0.5*m*v^2 = 0.5*k*0.08^2 =>K = 472.10625

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