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Butoxors [25]
2 years ago
7

A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t

he instant she leaves the cliff, determine the following.
Her gravitational potential energy relative to the water surface when she leaves the cliff
Her kinetic energy when she leaves the cliff
Her total mechanical energy relative to the water surface when she leaves the cliff
Her total mechanical energy relative to the water surface just before she enters the water.
The speed at which she enters the water.
Physics
1 answer:
Reil [10]2 years ago
7 0

Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

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The distance an object falls, from rest, in gravity is

                         D  =  (1/2) (G) (T²)

                        'T' is the number seconds it falls.

In this problem,

                         0.92 meter = (1/2) (9.8) (T²)

Divide each side by  4.9 :   0.92 / 4.9 = T²

Take the square root
of each side:                          √(0.92/4.9) = T

                                                  0.433 sec = T    

The horizontal speed doesn't make a bit of difference in
how long it takes to reach the floor.  BUT ... if you want to
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that with the horizontal speed.

The pencil is in the air for  0.433 second.
In that time, it travels
                                   (0.433s) x (1.4 m/s) = 0.606 meter

from the edge of the table.
 
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2 years ago
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Answer: 7.734 m/s

Explanation:

We have the following data:

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x=10 m  The horizontal distance of the water ballon

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We need to find the initial velocity V_{o} at which the water ballon was thrown, and we can find it by the following equation:

x=V_{o}cos \theta T (1)

Where T=2t is the total time the water ballon is on air

On the other hand, when we talk about parabolic motion (as in this situation) the water ballon reaches its maximum height just in the middle of this parabola, when V=0 and the time t is half the time T it takes the complete parabolic path.

So, if we use the following equation, we will find t:

V=V_{o}+gt=0 (2)

Isolating t:

t=\frac{-V_{o}}{g} (3)

Remembering T=2t:

T=2\frac{-V_{o}}{g} (4)

Substituting (4) in (1):

x=V_{o}cos \theta (2\frac{-V_{o}}{g}) (5)

Isolating V_{o}:

V_{o}=\sqrt{\frac{x g}{-2 cos \theta}} (6)

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