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ser-zykov [4K]
3 years ago
14

A basketball is thrown up into the air. It is released with an initial velocity of 8.5 m/s. How long does it take to get to the

top of its motion?
Physics
2 answers:
kompoz [17]3 years ago
7 0

Answer:

0.87 s

Explanation:

initial velocity, u = 8.5 m/s

Let it takes time t to reach to maximum height. At maximum height the velocity is zero, so, v = 0

Use first equation of motion

v = u - gt

where, g be the acceleration due to gravity

0 = 8.5 - 9.8 t

t = 0.87 s

Thus, the time taken to reach at top is 0.87 s.

ehidna [41]3 years ago
4 0
<h2>It takes 0.867 seconds to get to the top of its motion</h2>

Explanation:

We have equation of motion v = u + at

     Initial velocity, u = 8.5 m/s

     Final velocity, v = 0 m/s    - At maximum height

     Time, t = ?

     Acceleration , a = -9.81 m/s²

     Substituting

                      v = u + at  

                      0 = 8.5 + -9.81 x t

                      t = 0.867 s

  It takes 0.867 seconds to get to the top of its motion

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Given:-

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We know,

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Ans of this question A test charge of 1 couloumb moved from 30cm against the field of intensity 50N/c find the energy store in i
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A. Zero

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Given data,

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                                     = 9 x 10⁹ Nm²C⁻²

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Substituting the given values in the above equation

                            V₁ = 9 x 10⁹ x 30 / 0.3

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                            V₂ = 1.5 x 10¹² J

The energy stored in it is,

                             W = V₂ - V₁

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Hence, the energy stored in the charge is, W = 0        

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