A spring gun (k = 26 N/m) is used to shoot a 56-g ball horizontally. Initially the spring is compressed by 17 cm. The ball loses contact with the spring and leaves the gun when the spring is still compressed by 12 cm. What is the speed of the ball when it hits the ground, 1.3 m below the spring gun?

m/s

To find the speed of the ball when it hits the ground, we can use the principle of conservation of mechanical energy.

First, let's determine the potential energy stored in the compressed spring when it's compressed by 17 cm. The potential energy stored in a spring is given by the equation:

PE = (1/2) * k * x^2

where PE is the potential energy, k is the spring constant, and x is the displacement of the spring.

Plugging in the values for k and x, we get:

PE = (1/2) * 26 N/m * (0.17 m)^2
= 0.124 N*m

Now, let's find the gravitational potential energy of the ball when it hits the ground. The gravitational potential energy is given by the equation:

PE = m * g * h

where PE is the potential energy, m is the mass of the ball, g is the acceleration due to gravity, and h is the height.

Plugging in the values for m, g, and h, we have:

PE = 0.056 kg * 9.8 m/s^2 * 1.3 m
= 0.095 N*m

Since energy is conserved, the potential energy stored in the spring when compressed must be equal to the gravitational potential energy of the ball when it hits the ground.

0.124 N*m = 0.095 N*m

Now, let's find the kinetic energy of the ball when it hits the ground. The kinetic energy is given by the equation:

KE = (1/2) * m * v^2

where KE is the kinetic energy, m is the mass of the ball, and v is the velocity.

Since the potential energy is converted into kinetic energy when the ball leaves the gun, we can set the potential energy equal to the kinetic energy:

0.124 N*m = (1/2) * 0.056 kg * v^2

Solving for v, we have:

v^2 = (2 * 0.124 N*m) / 0.056 kg
= 4.429 m^2/s^2

v = sqrt(4.429 m^2/s^2)
= 2.10 m/s

Therefore, the speed of the ball when it hits the ground is approximately 2.10 m/s.