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A 107 gram apple falls from a branch that is 2 meters above the ground. (a) How much time elapses before the apple hits the ground

Sagot :

onyebuchinnaji

Answer:

The time of motion is 0.64 s.

Explanation:

Given;

mass of the apple, m = 107 g

height of fall, h = 2 m

The velocity of the apple when it hits the ground is calculated from the law of conservation of energy;

[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\gh = \frac{1}{2} v^2\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8\times 2} \\\\v = 6.261 \ m / s[/tex]

The time of motion is calculated;

v = u + gt

6.261 = 0 + 9.8t

6.261 = 9.8t

t = 6.261 / 9.8

t = 0.64 s

Therefore, the time of motion is 0.64 s

asuwafohjames

The time taken for the apple to hit the ground is 0.64 s.

The time taken for the apple to hit the ground can be calculated using the formula below.

Formula:

  • s = ut+gt²/2............ Equation 1

Where:

  • s = height
  • t = time
  • u = initial velocity
  • g = acceleration due to gravity.

 

From the question,

Given:

  • s = 2 m
  • u = 0 m/s (fall from a height)
  • g = 9.8 m/s²

Substitute these values into equation 1

  • 2 = 0(t)+9.8(t²)/2

Solve for t.

  • 9.8t² = 4
  • t² = 4/9.8
  • t² = 0.4081
  • t = √0.4081
  • t = 0.64 s.

Hence, The time taken for the apple to hit the ground is 0.64 s

Learn more about time taken here: https://brainly.com/question/4931057

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