A small rock is thrown vertically upward with a speed of 13.0 from the edge of the roof of a 31.0 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?

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To find the speed of the rock just before it hits the street, we need to consider the vertical motion of the rock. Let's break down the problem step by step:

Step 1: Calculate the initial velocity (upward)
Given:
Initial velocity (upward) = 13.0 m/s

Step 2: Calculate the final velocity (downward)
When the rock reaches the top of its trajectory, its velocity is zero. Therefore, the final velocity (downward) just before it hits the street will be the same as the initial velocity (upward), but with a negative sign.

Final velocity (downward) = -13.0 m/s

Step 3: Calculate the acceleration (due to gravity)
The only force acting on the rock throughout its trajectory is the force of gravity, which causes it to accelerate downward. The acceleration due to gravity near the Earth's surface is approximately 9.8 m/s^2, and we assume air resistance is negligible.

Acceleration (due to gravity) = -9.8 m/s^2

Step 4: Apply the kinematic equation
We can use the kinematic equation:
v^2 = u^2 + 2as

where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement

In this case, we want to find the final velocity (v) when the rock hits the street. The initial velocity (u) is given as 13.0 m/s, the acceleration (a) is -9.8 m/s^2, and the displacement (s) is the height of the building, 31.0 meters.

Applying the equation:
v^2 = u^2 + 2as

v^2 = (13.0 m/s)^2 + 2(-9.8 m/s^2)(-31.0 m)
v^2 = 169.0 m^2/s^2 + 6076 m^2/s^2
v^2 = 6245 m^2/s^2

Taking the square root of both sides:
v = √6245 m/s
v ≈ 79.0 m/s

Therefore, the speed of the rock just before it hits the street is approximately 79.0 m/s.

To calculate the time elapsed from when the rock is thrown until it hits the street, we can use the equation of motion:

s = ut + 0.5at^2

where:
s = displacement (31.0 m)
u = initial velocity (13.0 m/s)
a = acceleration (due to gravity, -9.8 m/s^2)
t = time

Rearranging the equation:
0.5at^2 + ut - s = 0

Substituting the known values:
0.5(-9.8)t^2 + (13.0)t - 31.0 = 0

This is a quadratic equation, so we can solve it using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 0.5(-9.8), b = 13.0, and c = -31.0.

Calculating:
t = [ -13.0 ± √(13.0^2 - 4(0.5)(-9.8)(-31.0)) ] / (2(0.5)(-9.8))
t = [ -13.0 ± √(169 + 6076) ] / (-9.8)
t = [ -13.0 ± √(6245) ] / (-9.8)

Using the positive value from the square root for time (since time cannot be negative in this context):

t = [ -13.0 + √(6245) ] / (-9.8)
t ≈ 2.0 s

Therefore, the time elapsed from when the rock is thrown until it hits the street is approximately 2.0 seconds.