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A quarter is flipped from a height of 1.45 m above the ground. How much time will it take to reach the ground if the person flipping the coin gave it a velocity of 10.32 m/s? Please answer in seconds and to the third decimal place.

Sagot :

Lanuel

It will take the quarter 0.151 seconds to reach the ground.

Given the following data:

  • Height = 1.45 meters
  • Initial velocity = 10.32 m/s

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find how much time it will take the quarter to reach the ground, we  would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

[tex]S = ut + \frac{1}{2} at^2[/tex]

Where:

  • S is the height or distance covered.
  • u is the initial velocity.
  • a is the acceleration.
  • t is the time measured in seconds.

Substituting the values into the formula, we have;

[tex]1.45 = 10.32(t) + \frac{1}{2} (9.8)t^2\\\\1.45 = 10.32t + 4.9t^2\\\\4.9t^2 + 10.32t - 1.45 = 0[/tex]

The standard form of a quadratic equation is:

[tex]ax^2 + bx + c = 0[/tex]

a = 4.9, b = 10.32 and c = 1.45

We would solve the above quadratic equation by using the quadratic equation formula;

[tex]x = \frac{-b\; \pm \;\sqrt{b^2 - 4ac}}{2a}[/tex]

Substituting the values, we have;

[tex]t = \frac{-10.32\; \pm \;\sqrt{10.32^2\; - \;4(4.9)(1.45)}}{2(4.9)}\\\\t = \frac{-10.32\; \pm \;\sqrt{106.5024\; - \;28.42}}{9.8}\\\\t = \frac{-10.32\; \pm \;\sqrt{78.0824}}{9.8}\\\\t = \frac{-10.32\; \pm \;8.84}{9.8}\\\\t = \frac{-10.32\; + \;8.84}{9.8}\\\\t = \frac{1.48}{9.8}[/tex]

Time, t = 0.151 seconds.

Therefore, it will take the quarter 0.151 seconds to reach the ground.

Read more: https://brainly.com/question/8898885

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