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A person throws an object downwards from the top of a building. The building is 37.9 m high. The object hits the ground at the base of the building with a velocity of 28.6 m/s.

Calculate the time taken between the object being thrown and hitting the ground.

Answer :

Final answer:

The time taken for the object to hit the ground can be calculated using the equation d = v0t + (1/2)at^2. Substituting the given values into the equation, the time taken is approximately 7.7 seconds.

Explanation:

The time taken for the object to hit the ground can be calculated using the equation:

d = v0t + (1/2)at2

Where:
d is the distance traveled (37.9 m)
v0 is the initial velocity (0 m/s)
a is the acceleration due to gravity (-9.8 m/s2)
t is the time taken

Substituting the given values into the equation, we have:

37.9 = (0)t + (1/2)(-9.8)t2

Simplifying the equation and solving for t, we get:

t2 - 7.7t = 0

Factoring out t, we get:

t(t - 7.7) = 0

This gives us two possible solutions: t = 0 and t = 7.7s

Since the object is thrown downwards and hits the ground, the time taken can only be positive. Therefore, the time taken between the object being thrown and hitting the ground is approximately 7.7 seconds.

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