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2 years ago

5

A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above

the ground. How much time elapses between the instant of release and the instant of impact with the ground

1 answer:

erastova [34]2 years ago

6 0

Answer:

t = 2 s

Explanation:

In order to find the time taken by the stone to fall from the top of the building to the ground we can use 2nd equation of motion. 2nd equation of motion is as follows:

s = Vit + (0.5)gt²

where,

t = time = ?

Vi = Initial Velocity = 20 m/s

s = height of building = 60 m

g = 9.8 m/s²

Therefore,

60 m = (20 m/s)t + (0.5)(9.8 m/s²)t²

4.9t² + 20t - 60 = 0

solving this quadratic equation we get:

t = -6.1 s (OR) t = 2 s

Since, the time cannot be negative in magnitude.

Therefore,

<u>t = 2 s</u>

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