A small rock is thrown vertically upward with a speed of 14.0m/s from the edge of the roof of a 25.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.

What is the speed of the rock just before it hits the street?

How much time elapses from when the rock is thrown until it hits the street?

Thank you

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To find the answers to these questions, we can use the equations of motion for freefall.

First, let's find the time it takes for the rock to reach its maximum height. We can use the equation:

v = u + at

Where:
- v is the final velocity (0 since the rock reaches its highest point and momentarily comes to rest),
- u is the initial velocity (14.0 m/s, since the rock is thrown vertically upward),
- a is the acceleration due to gravity (-9.8 m/s^2, assuming down is the positive direction),
- t is the time.

Rearranging the equation, we get:

t = (v - u) / a

Substituting the values, we find:

t = (0 - 14.0) / -9.8
t = 1.43 seconds (rounded to two decimal places)

Now, let's find the time it takes for the rock to fall back down from its maximum height to the street. We can use the same equation, but this time the initial velocity is 0 since the rock starts from rest:

t = (v - u) / a

Substituting the values, we find:

t = (v - 0) / -9.8
t = -v / 9.8

The negative sign indicates that we are considering the downward direction. Solving for t, we get:

-1.43 = v / 9.8

v = -1.43 x 9.8
v = -14 m/s

Since we are only interested in the magnitude of the velocity, we take the absolute value:

v = 14 m/s

Therefore, the speed of the rock just before it hits the street is 14 m/s.

Now let's find the total time of flight, which is the sum of the time it takes to reach the maximum height and the time it takes to fall back down:

Total time = 1.43 + 1.43 = 2.86 seconds.

So, it takes approximately 2.86 seconds for the rock to hit the street after it is thrown.