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AysviL
9 days ago
14

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 th

e ground? (Disregard air resistance. a = −g = −9.81 m/s 2 .)
Physics
1 answer:
Maru [2.3K]9 days ago
6 0

Starting speed of the coin (u)= 0 (Since it starts from rest)

Gravitational acceleration (a) = g = 9.81 m/s²

Duration of the fall (t) = 1.5 s

According to the motion equation we can state:

\boxed{ \bf{v = u + at}}

Substituting the known values into the equation yields:

\longrightarrow v = 0 + 9.81 × 1.5

\longrightarrow v = 14.715 m/s

\therefore The speed of the coin upon reaching the ground/Final speed of the coin = 14.715 m/s

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1 month ago
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A small block of mass 200 g starts at rest at A, slides to B where its speed is vB=8.0m/s,vB=8.0m/s, then slides along the horiz
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Answer

Data provided:

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kinetic energy calculated as

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b) using the formula v² = u² + 2 a s

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28 days ago
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