well, the Potential energy in the spring must equal the change in PE of the projectile.
1/2 k x^2=mgh
solve for k
Then speed of the projectile at equilibrum
1/2 kx^2=1/2 m v^2 solve for v.
If the spring is compressed a distance of 0.144 m and the gun fired vertically as shown, the gun can launch a 18.5g projectile from rest to a maximum height of 19.5 m above the starting point of the projectile. Neglecting all resistive forces, determine the spring constant
Determine the speed of the projectile as it moves through the equilibrium position of the spring (where x = 0)
1 answer
1 answer