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erica [24]
3 years ago
14

**URGENT** Now lets look at a skydiver who jumps out of a plane that has a forward velocity of 40 m/s. ignore air resistance. Af

ter 1 second, what is the skydivers downward velocity?
Physics
1 answer:
Dennis_Churaev [7]3 years ago
5 0
If there's no air resistance, then the diver keeps moving horizontally at 40 m/s, even after s/he jumps out of a perfectly good airplane.

Also, gravity ADDS 9.8 m/s of downward speed for every second the diver falls.

One second after leaving the airplane, the diver is moving horizontally at 40 m/s, and moving vertically downward at 9.8 m/s .

The question asks for the "downward velocity".  That's 9.8 m/s .


The diver's TOTAL velocity is

                       √ (40² + 9.8²)

                   =  √ (1,600 m²/s² + 96.04 m²/s²)

                   =  √ (1,696.04 m²/s²)

                   =        41.2 m/s  in some direction between
                                           straight forward and straight down.
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The work required is -515,872.5 J

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Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

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W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

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