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If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance, after 10 s: How fast (speed) would it be traveling (in m/s)

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Answer:

Step-by-step explanation:

Using the equation of motion v = u + at to get the speed at which the object would be travelling.

v is the final speed (in m/s)

u is the initial velocity (in m/s)

a is the acceleration (in m/s²)

t is the time taken (in secs)

Given parameters

u = 0m/s

t = 10s

a = g = 9.8m/s²

Substituting this values into the formula;

v = 0+9.8(10)

v = 0+ 98

v = 98m/s

Hence the rock will be travelling at a speed of 98m/s.

User RobinCominotto
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