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A coin released at rest from the top of a tower hits the ground after falling 1.5 s. What is the speed of the coin as it hits the ground? (Disregard air resistance. a = −g = −9.81 m/s 2 .)

Respuesta :

Cathenna

Initial speed of the coin (u)= 0 (As the coin is released from rest)

Acceleration due to gravity (a) = g = 9.81 m/s²

Time of fall (t) = 1.5 s

From equation of motion we have:

[tex] \boxed{ \bf{v = u + at}}[/tex]

By substituting values in the equation, we get:

[tex] \longrightarrow [/tex] v = 0 + 9.81 × 1.5

[tex] \longrightarrow [/tex] v = 14.715 m/s

[tex] \therefore [/tex] Speed of the coin as it hits the ground/Final speed of the coin = 14.715 m/s

asuwafohjames

The speed of the coin as it hits the ground is 14.715 m/s.

We can solve the problem above using the equation of acceleration under gravity.

⇒ Equation:

v = u+gt................... Equation 1

⇒ Where:

  • The final velocity of the coin
  • u = Initial velocity of the coin
  • g = acceleration due to gravity
  • t = time

From the question,

⇒ Given:

  • u = 0 m/s (from rest)
  • t = 1.5 s
  • g = -9.81 m/s²

Substitute these values into equation 1

  • v = 0+1.5(-9.81)
  • v = -14.715 m/s

Note the speed has a negative sign because it acts in the same direction as the acceleration due to gravity

Hence, the speed of the coin as it hits the ground is 14.715 m/s.

Learn more about gravitational motion here: https://brainly.com/question/17785701

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