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sweet [91]
3 years ago
6

A basketball player can leap upward 0.43 m. how long does he remain in the air? use an acceleration due to gravity of 9.80 m/s2

and find your answer to the nearest 0.001 s.
Physics
1 answer:
MArishka [77]3 years ago
3 0
From the equations of linear motion,
v² = u² + 2as where v is the final velocity, u is the initial velocity and a is the gravitational acceleration, and s is the displacement,
Thus, v² = u² -2gs, but v=0
hence, u² = 2gs
                = 2×9.81×0.43
                = 8.4366
            u = √8.4366
               =2.905 m/s
Hence the initial velocity is 2.905 m/s
 Then using the equation v= u +gt .
Therefore, v = u -gt. (-g because the player is jumping against the gravity)
but, v = 0
Thus, u= gt
Hence, t = u/g
              = 2.905/9.81
              = 0.296 seconds


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The given parameters;

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The acceleration due to gravity is calculated as follows;

h = ut - \frac{1}{2} gt^2\\\\100 = 15(10) - (0.5\times 10^2)g\\\\100 = 150 - 50g\\\\50g = 150-100\\\\50g = 50\\\\g = 1 \ m/s^2

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2 years ago
A student makes a short electromagnet by winding 300 turns of wire around a wooden cylinder of diameter d 5.0 cm. The coil is co
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A is area

We are given

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Let's make z the subject ;

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