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____ [38]
2 years ago
7

A juggler is practicing with one ball. It takes 2.4 seconds for the ball to leave her hand and return to her hand. How fast did

the juggler initially throw the ball upwards?
Physics
1 answer:
nikklg [1K]2 years ago
5 0

The kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s

Given parameters

  • Time t = 2.4 s

To find

  • Initial velocity

Kinematics is the part of physics that establishes the relationships between the position, velocity, and acceleration of bodies.

In this case we have a vertical launch

          y = y₀ + v₀ t - ½ g t²

Where y and y₀ are the final and initial positions, respectively, v₀ the initial velocity, g the acceleration of gravity (g = 9.8 m / s²) and t the time

   

With the ball in hand, its position is zero

         0 = 0 + v₀ t - ½ g t²

         v₀ t - ½ g t² = 0

         v₀ = ½ g t

Let's calculate

         v₀ = ½ 9.8 2.4

         v₀ = 11.76 m / s

In conclusion using kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s

Learn more about vertical launch kinematics here:

brainly.com/question/15068914

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Answer:

  t = 0.714 s and  x = 5.0 m

Explanation:

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         vₓ = 7.0 m / s

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         y = y₀ + v_{oy} t - ½ g t²

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        0 = y₀ + 0 - ½ g t²

        t = \sqrt{\frac{2y_o}{g}  }

        t = ra 2 2.5 / 9.8

        t = 0.714 s

the distance traveled is

       x = vₓ t

       x = 7.0 0.714

       x = 5.0 m

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