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egoroff_w [7]
3 years ago
6

In getting ready to slam-dunk the ball, a basketball player starts from rest and sprints to a speed of 5.45 m/s in 3.02 s. Assum

ing that the player accelerates uniformly, determine the distance he runs.
Physics
1 answer:
vova2212 [387]3 years ago
8 0

Recall that average velocity <em>v</em> is given by

<em>v</em> = ∆<em>x</em>/∆<em>t</em>

where ∆<em>x</em> is displacement and ∆<em>t</em> is time.

Under constant acceleration, average velocity is also equal to the average of the initial and final velocities,

<em>v</em> = (<em>v</em>₂ + <em>v</em>₁)/2

The player starts at rest, so <em>v</em>₁ = 0, and speeds up to <em>v</em>₂ = 5.45 m/s in a matter of ∆<em>t</em> = 3.02 s. So

∆<em>x</em> = (<em>v</em>₂ + <em>v</em>₁) ∆<em>t</em> / 2

∆<em>x</em> = (5.45 m/s) * (3.02 s) / 2

∆<em>x</em> ≈ 8.23 m

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Given gravitational potential energy when he's lifted is 2058 J.

Kinetic energy is transferred to the person.

Amount of kinetic energy the person has is -2058 J

velocity of person = 7.67 m/s².

<h3>Explanation:</h3>

Given:

Weight of person = 70 kg

Lifted height = 3 m

1. Gravitational potential energy of a lifted person is equal to the work done.

PE_g=W=m\times g\times h\\Acceleration due to gravity = g = 9.8 \ m/s^2 \\PE_g= m = m\times g\times h= 70\times 9.8 \times 3 = 2058\ kg.m/s^2 = 2058\ J

Gravitational potential energy is equal to 2058 Joules.

2. The Gravitational potential energy is converted into kinetic energy. Kinetic energy is being transferred to the person.

3. Kinetic energy gained = Potential energy lost = -PE_g = -2058\ kg.m/s^2

Kinetic energy gained by the person = (-2058 kg.m/s²)

4. Velocity = ?

Kinetic energy magnitude= \frac{1}{2} m\times v^2 = m\times g \times h

Solving for v, we get

v=\sqrt{2gh} =\sqrt{2\times 9.8 \times 3} = \sqrt{58.8} = 7.67 m/s^2

The person will be going at a speed of 7.67 m/s².

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