A spring gun (k = 27.1 N/m) is used to shoot a 51.1-g ball horizontally. Initially the spring is compressed by 16.5 cm. The ball loses contact with the spring and leaves the gun when the spring is still compressed by 11.5 cm. What is the speed of the ball when it hits the ground, 1.41 m below the spring g

kx₁²/2 -kx₂²/2 = mv(x)²/2

Find v²(x)
Vertical motion:
h=gt²/2 => t=sqrt(2h/g)
v(y)=gt.
v=sqrt{v²(x)+ v²(y)}

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 before the ball is released. The potential energy stored in a spring when it is compressed or stretched can be calculated using the formula:

Potential Energy = (1/2) k x^2

where k is the spring constant and x is the displacement from the equilibrium position.

Given that the spring constant, k, is 27.1 N/m and the displacement, x, is 16.5 cm (or 0.165 m), we can calculate the potential energy stored in the spring:

Potential Energy = (1/2) * 27.1 * (0.165)^2 = 0.564 J

Since energy is conserved, this potential energy is transferred into kinetic energy when the ball is released. The kinetic energy of an object can be calculated using the formula:

Kinetic Energy = (1/2) * m * v^2

where m is the mass of the object and v is its velocity.

In this case, the mass of the ball is 51.1 g (or 0.0511 kg). We need to find the velocity, v, when the ball hits the ground, which is 1.41 m below the spring. This means that the ball has fallen a distance of 1.41 m.

Using the principle of conservation of energy, we can set the potential energy equal to the kinetic energy:

Potential Energy = Kinetic Energy

0.564 J = (1/2) * 0.0511 kg * v^2

Now, let's solve for v:

v^2 = (2 * Potential Energy) / m
v^2 = (2 * 0.564 J) / 0.0511 kg
v^2 = 22.08 m^2/s^2
v = √22.08 m/s

Therefore, the speed of the ball when it hits the ground, 1.41 m below the spring, is approximately 4.70 m/s.