# A ball with a mass of #300 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #12 (kg)/s^2# and was compressed by #7/6 m# when the ball was released. How high will the ball go?

Maximum altitude

Ignoring air resistance, other sources of friction, etc., this problem can be solved with energy conservation:

Initially, assuming the spring is compressed to point where its height is insignificant (so that we need not worry about gravitational potential energy of the ball beforehand), all energy is stored as spring potential energy in the spring. Because nothing is moving prior to the launch, there is no kinetic energy. Therefore, we have only potential energy to begin with.

After the launch, as a the spring decompresses, the energy stored within it as spring potential energy is transferred to the ball, which is observed as the ball shoots upward. The spring potential energy has been transformed into kinetic energy. However, as the ball rises, it is gaining altitude, and therefore gaining gravitational potential energy. Because the potential energy of the ball increases, its kinetic energy must decrease, and we observe this in the form of the ball stopping at some point in the air before falling back to the earth. This is the point at which all of the kinetic energy has been transformed into gravitational potential energy.

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To find the maximum height the ball will reach, you can use the principle of conservation of mechanical energy. The potential energy stored in the spring when compressed is converted into kinetic energy of the ball when released, and then into gravitational potential energy at the maximum height.

The potential energy stored in the compressed spring is given by ( \frac{1}{2} kx^2 ), where ( k ) is the spring constant and ( x ) is the compression distance.

Given:

- Mass of the ball (( m )): 300 g = 0.3 kg
- Spring constant (( k )): 12 (kg/s^2)
- Compression distance (( x )): ( \frac{7}{6} ) m

The potential energy stored in the spring is: [ \text{Potential energy} = \frac{1}{2} kx^2 ]

Then, equate this to the gravitational potential energy at the maximum height: [ mgh = \frac{1}{2} kx^2 ]

Solve for ( h ), which represents the maximum height the ball reaches.

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To find the maximum height the ball will reach, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the compressed spring is equal to the final kinetic energy of the ball at its maximum height.

The potential energy stored in the compressed spring is given by:

[ PE = \frac{1}{2}kx^2 ]

where ( k ) is the spring constant and ( x ) is the compression of the spring. Plugging in the values:

[ PE = \frac{1}{2} \times 12 \times \left( \frac{7}{6} \right)^2 ]

[ PE = \frac{1}{2} \times 12 \times \frac{49}{36} ]

[ PE = 7 , \text{J} ]

At maximum height, all the initial kinetic energy of the ball is converted into potential energy. The initial kinetic energy of the ball is given by:

[ KE = \frac{1}{2}mv^2 ]

where ( m ) is the mass of the ball and ( v ) is its velocity at the release point. Since the ball is projected vertically, its initial velocity is 0.

Therefore, the initial kinetic energy is also 0.

At the maximum height, the potential energy equals the initial potential energy:

[ PE = mgh ]

where ( m ) is the mass of the ball, ( g ) is the acceleration due to gravity, and ( h ) is the maximum height.

Setting ( PE ) equal to ( mgh ) and solving for ( h ):

[ mgh = 7 ]

[ 0.3 \times 9.8 \times h = 7 ]

[ h = \frac{7}{0.3 \times 9.8} ]

[ h ≈ 2.38 , \text{m} ]

So, the ball will reach a maximum height of approximately 2.38 meters.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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