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Physics / College Physics 1 / Chapter 4 / Problem 46

College Physics | 1st Edition | ISBN: 9781938168000 | Authors: Paul Peter Urone, Roger Hinrichs

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Integrated Concepts A basketball player jumps straight up

Chapter 4, Problem 46

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QUESTION:

Problem 46PE

Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.

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QUESTION:

Problem 46PE

Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.

ANSWER:

Solution 46PE

The basketball player wants to reach a given height. From this, we have to calculate the velocity when he jumps from the floor. From this data, we can further calculate the acceleration. We are also required to calculate the force exerted by him on the ground.

Part a

Step 1 of 3

At the maximum height, his final velocity $v$ would be zero.

Let the initial velocity be $u$ .

Height to reach $h=0.900\$ m

From the equation,

${v}^{2}={u}^{2}-2gh$ , (negative symbol as he moves up)

${0}^{2}={u}^{2}-2\times{}9.8$ m/s2 $\times{}0.900$ m

$u=\sqrt{2\times{}9.8\times{}0.900}$ m/s

$u=4.20$ m/s

Therefore, his velocity while leaving the floor will be 4.20 m/s.

Part b

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