A ball with a mass of #480 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #16 (kg)/s^2# and was compressed by #3/5 m# when the ball was released. How high will the ball go?

Answer 1

The height is #=0.61m#

The spring constant is #k=16kgs^-2#

The compression is #x=3/5m#

The potential energy is

#PE=1/2*16*(3/5)^2=2.88J#

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 #

The sacceleration due to gravity is #g=9.8ms^-2#

Then ,

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

Mass of the ball is #m=0.48kg#

#PE_(ball)=5.76=0.48*9.8*h#

#h=2.88*1/(0.48*9.8)#

#=0.61m#

The height is #=0.61m#

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Answer 2

To find the maximum height, we can use the conservation of mechanical energy principle, where the initial mechanical energy (potential energy stored in the spring) equals the final mechanical energy (kinetic energy at maximum height).

The potential energy stored in the spring when compressed by ( \Delta x ) is given by ( PE = \frac{1}{2}k(\Delta x)^2 ).

The kinetic energy at maximum height is given by ( KE = mgh ), where ( m ) is the mass of the ball, ( g ) is the acceleration due to gravity (9.8 m/s(^2)), and ( h ) is the maximum height.

Equating the potential and kinetic energies:

[ \frac{1}{2}k(\Delta x)^2 = mgh ]

Rearranging for ( h ):

[ h = \frac{\frac{1}{2}k(\Delta x)^2}{mg} ]

Substituting the given values:

[ h = \frac{\frac{1}{2}(16)(\frac{3}{5})^2}{0.48 \times 9.8} ]

[ h \approx \frac{0.5 \times 16 \times \frac{9}{25}}{4.704} \approx \frac{72 \times \frac{9}{25}}{4.704} \approx \frac{648}{4.704} \approx 137.93 , \text{m} ]

So, the ball will reach a maximum height of approximately 137.93 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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