A ball with a mass of #120 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #96 (kg)/s^2# and was compressed by #4/3 m# when the ball was released. How high will the ball go?

Answer 1

The height is #=72.56m#

The spring constant is #k=96kgs^-2#

The compression is #x=4/3m#

The potential energy in the spring is

#PE=1/2*96*(4/3)^2=85.3J#

This potential energy will be converted to kinetic energy when the spring is released and to potential energy of the ball

#KE_(ball)=1/2m u^2#

Let the height of the ball be #=h #

Then ,

The potential energy of the ball is #PE_(ball)=mgh#

#PE_(ball)=85.3=0.12*9.8*h#

#h=85.3*1/(0.12*9.8)#

#=72.56m#

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Answer 2

To find the maximum height reached by the ball, we can use the principle of conservation of mechanical energy. At the highest point, all of the kinetic energy will be converted into potential energy. The initial potential energy stored in the spring is equal to the kinetic energy of the ball at its maximum height.

The potential energy stored in the spring, (U_s), is given by: [ U_s = \frac{1}{2} k x^2 ]

Where: ( k ) = spring constant = 96 (kg)/s^2 ( x ) = compression of the spring = 4/3 m

Substituting the values: [ U_s = \frac{1}{2} \times 96 \times \left(\frac{4}{3}\right)^2 ]

[ U_s = \frac{1}{2} \times 96 \times \frac{16}{9} ]

[ U_s = 64 ]

The potential energy at maximum height, (U_{max}), is equal to the potential energy stored in the spring: [ U_{max} = 64 ]

At maximum height, all of the kinetic energy is converted into potential energy. The kinetic energy of the ball, (K), is given by: [ K = \frac{1}{2} m v^2 ]

Where: ( m ) = mass of the ball = 120 g = 0.12 kg ( v ) = velocity of the ball at maximum height (which is 0 at maximum height)

Substituting the values: [ K = \frac{1}{2} \times 0.12 \times 0^2 ]

[ K = 0 ]

At maximum height, ( U_{max} = K ) [ 64 = 0 ]

Now, we can use the equation for potential energy at maximum height to find the height, (h), above the release point: [ U_{max} = mgh ]

Solving for (h): [ h = \frac{U_{max}}{mg} ] [ h = \frac{64}{0.12 \times 9.8} ] [ h ≈ 54.24 , \text{m} ]

Therefore, the maximum height reached by the ball is approximately 54.24 meters.

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Answer from HIX Tutor

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.

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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