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2 years ago

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A 0.0500-kg golf ball heads off from the tee with an initial speed of 78.2 m/s and reaches to its maximum height of 37.8 m. If a

ir resistance is neglected, (a) What is the kinetic energy of the ball at the pinnacle of its trajectory? (b) What is its speed when it is 5.60 m below the pinnacle point?

1 answer:

Marianna [84]2 years ago

3 0

(a) 134.4 J

The total mechanical energy of the ball at the moment of launch is just equal to its initial kinetic energy:

$E=K_i = \frac{1}{2}mu^2$

where

m = 0.05 kg is the mass of the ball

u = 78.2 m/s is the initial speed

Substituting,

$E=\frac{1}{2}(0.05)(78.2)^2=152.9 J$

At the pinnacle of its trajectory, the total mechanical energy is sum of kinetic energy and potential energy:

$E=K_f + U_f = K_f + mgh$

where

$g=9.8 m/s^2$ is the acceleration of gravity

h = 37.8 m is the maximum height

Since the total energy must be conserved,

E = 152.9 J

Therefore, we can solve to find the kinetic energy of the ball at the pinnacle:

$K_f = E-mgh=152.9-(0.05)(9.8)(37.8)=134.4 J$

b) 74.2 m/s

When the ball is 5.60 m below the pinnacle point, the heigth of the ball is

$h=37.8-5.6=32.2 m$

So its potential energy is

$U=mgh=(0.05)(9.8)(32.2)=15.8 J$

The total energy is again the sum of potential and kinetic energy:

E = K + U

So the kinetic energy at that point is

$K=E-U=152.9-15.8=137.1 J$

And since the kinetic energy is

$K=\frac{1}{2}mv^2$

We can find the speed:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(137.1)}{0.05}}=74.2 m/s$

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The question is missing some parts. Here is the complete question.

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Calculating velocity for each temperature:

For 300K:

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For 900K:

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