# A rock is thrown downward into a well that is 8.58 m deep. If the splash is heard 1.08 seconds later, what was the initial speed of the rock?

## time for sound to come up the well:

time=depth/speedsound=8.58/342m/s

subtract that from 1.08 seconds.
The remainder is the time it took the rock to get to the bottom.

depth=vi*t+1/2 g t^2
put in time, depth, and g, solve for vi.

## Well, that's quite a "rock"-ing question! To solve this, we can break it down into a few steps.

First, we need to find the time it takes for the rock to reach the bottom of the well. Since it's thrown downward, we know it's being affected by gravity. Using a formula, we find that it takes approximately 1.08 seconds to hit the bottom.

Now, let's assume that the initial speed of the rock is "V" (which stands for "Very fast rock"). As it falls, gravity will accelerate it downwards. We can use another formula to find the acceleration due to gravity (9.8 m/s²) multiplied by the time it takes (1.08 seconds). This gives us the change in speed.

Finally, we can use the equation of motion to find the initial speed. The initial velocity (the speed at which it was thrown downwards) plus the change in speed due to gravity must add up to zero when it hits the bottom.

So, using this hilarious formulation, we can solve the equation by putting a smile on our faces and using the formula V + (-9.8 m/s² * 1.08 seconds) = 0.

Solving for V, we have V = 9.8 m/s² * 1.08 seconds.

Hence, the initial velocity (speed) of the rock is approximately 10.58 m/s. Keep on rockin', my friend!

## To find the initial speed of the rock, we can use the equation of motion for objects in free fall.

The equation is given by:

y = ut + (1/2)gt^2

where:
y = vertical distance traveled (in meters)
u = initial velocity (in m/s)
t = time taken (in seconds)
g = acceleration due to gravity (approximately 9.8 m/s^2)

In this case, the initial position of the rock is at the top of the well (y = 0) and the final position is at the bottom of the well (y = -8.58 m).

Using the equation, we can rearrange it to solve for the initial velocity (u):

u = (y - (1/2)gt^2) / t

Substituting the given values:

u = (-8.58 m - (1/2)(9.8 m/s^2)(1.08 s)^2) / 1.08 s

Calculating the equation:

u = (-8.58 m - (1/2)(9.8 m/s^2)(1.1664 s^2)) / 1.08 s
u = (-8.58 m - 6.0032 m) / 1.08 s
u = -14.5832 m / 1.08 s
u ≈ -13.50 m/s

However, speed cannot be negative in this context, so we take the absolute value of the result:

Initial speed ≈ 13.50 m/s

Therefore, the initial speed of the rock was approximately 13.50 m/s.

## To determine the initial speed of the rock, we can use the kinematic equation for vertical motion:

h = (1/2)gt^2 + v0t + h0

Where:
h is the height of the well (8.58 m)
g is the acceleration due to gravity (-9.8 m/s^2)
t is the time (1.08 s)
v0 is the initial velocity (unknown)
h0 is the initial height (0 m, since we are measuring from the top of the well)

Rearranging the equation, we can solve for v0:

v0 = (h - (1/2)gt^2) / t

Substituting the given values into the equation, we have:

v0 = (8.58 m - (1/2)(-9.8 m/s^2)(1.08 s)^2) / 1.08 s

Simplifying the expression inside the parentheses:

v0 = (8.58 m - (1/2)(-9.8 m/s^2)(1.1664 s^2)) / 1.08 s

v0 = (8.58 m - (-5.456 m)) / 1.08 s

v0 = (8.58 m + 5.456 m) / 1.08 s

v0 = 13.036 m / 1.08 s

v0 ≈ 12.07 m/s

Therefore, the initial speed of the rock was approximately 12.07 m/s.