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 .)

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:

From the question,

⇒ Given:

Substitute these values into equation 1

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: link

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:

By substituting values in the equation, we get:

v = 0 + 9.81 × 1.5

v = 14.715 m/s

Speed of the coin as it hits the ground/Final speed of the coin = 14.715 m/s

v=ir = 3x15 =45 v

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