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Snezhnost [94]
3 years ago
6

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.260 m and then accelerates through this dista

nce by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.920 m above the floor. (a) Calculate his velocity (in m/s) when he leaves the floor.
Physics
1 answer:
Stels [109]3 years ago
3 0

Answer:

u = 4.25 m/s

Explanation:

The kinematic expression is as follows:

v² = u² - 2as

where

u = initial velocity

v = final velocity

s = distance

g = acceleration due to gravity .

when he reaches a height of 0.920 m above the floor the final velocity  = 0

acceleration due to gravity act opposite the initial direction of motion. So, -9.81 m/s.

v² = u² + 2as

0² - u² = 2 (- 9.81) × 0.920

- u² = 2 × - 9.81 × 0.920

multiply both sides by -1

u² = 2 × 9.81 × 0.920

u² = 18.0504

u = √18.0504

u = 4.2485762321

u = 4.25 m/s

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Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

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The equation of kinetic energy is the following:

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Then:

-(final PE - initial PE) = final KE - initial KE          

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The speed of the ball 1.0 m above the ground is 44 m/s.

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