A ball with a mass of #240 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 #1/2 m# when the ball was released. How high will the ball go?

Answer 1

The height is #=0.64m#

The spring constant is #k=12kgs^-2#

The compression is #x=1/2m#

The potential energy is

#PE=1/2*12*(1/2)^2=1.5J#

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

#KE=1/2m u^2#

The initial velocity is #=u#

#u^2=2/m*KE=2/m*PE#

#u^2=2/0.24*1.5=12.5#

#u=sqrt12.5=3.54ms^-1#

Resolving in the vertical direction #uarr^+#

We apply the equation of motion

#v^2=u^2+2ah#

At the greatest height, #v=0#

and #a=-g#

So,

#0=12.5-2*9.8*h#

#h=12.5/(2*9.8)=0.64m#

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Answer 2

To find the maximum height ( h ) reached by the ball, we can use the principle of conservation of mechanical energy:

[ PE_{initial} + KE_{initial} = PE_{final} + KE_{final} ]

where ( PE ) is the potential energy and ( KE ) is the kinetic energy.

At the point of release, all the energy is potential energy stored in the spring:

[ PE_{initial} = \frac{1}{2} k x^2 ]

where ( k ) is the spring constant and ( x ) is the compression of the spring.

At maximum height, all the energy is potential energy:

[ PE_{final} = mgh ]

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

Since the ball is projected vertically, the initial kinetic energy is zero.

[ KE_{initial} = 0 ]

At maximum height, the final kinetic energy is also zero.

[ KE_{final} = 0 ]

So, the equation becomes:

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

Solve for ( h ):

[ h = \frac{1}{2} \times \frac{k x^2}{mg} ]

Given that ( m = 240 ) g, ( k = 12 ) (kg)/s(^2), ( x = \frac{1}{2} ) m, and ( g = 9.8 ) m/s(^2), we can substitute these values into the equation to find the height ( h ):

[ h = \frac{1}{2} \times \frac{12 \times (\frac{1}{2})^2}{0.240 \times 9.8} ]

[ h = \frac{1}{2} \times \frac{12 \times \frac{1}{4}}{2.352} ]

[ h = \frac{1}{2} \times \frac{3}{2.352} ]

[ h ≈ \frac{3}{4.704} ]

[ h ≈ 0.6375 , \text{m} ]

Therefore, the ball will reach a height of approximately 0.6375 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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