1) A penny is dropped from rest from the top of a very tall building. Assuming the height of the building is 407 m and ignoring air resistance, find the speed with which the penny strikes the ground.

I am not sure how to start this. Thank you!

V = sqrt(2 g X)

Why? Because
M g X = (1/2) M V^2

To find the speed with which the penny strikes the ground, we can use the principles of kinematics and the equations of motion.

First, let's define our variables:
- Initial velocity (u) = 0 m/s (since the penny is dropped from rest)
- Final velocity (v) = ?
- Acceleration (a) = 9.8 m/s² (acceleration due to gravity, assuming downward direction)
- Time taken (t) = ?
- Distance (s) = 407 m

We can use the equation of motion that relates distance, acceleration, time, and initial velocity:

s = ut + 1/2at²

In this case, the initial velocity is zero (u = 0), so the equation simplifies to:

s = 1/2at²

Now we can substitute the known values into the equation and solve for time (t):

407 = 0.5 * 9.8 * t²

407 = 4.9t²

Divide both sides by 4.9:

407/4.9 = t²

83.06 = t²

Now take the square root of both sides to find t:

t ≈ √83.06
t ≈ 9.11 seconds (rounded to two decimal places)

Now that we have the time it takes for the penny to fall, we can find the final velocity (v) using the equation:

v = u + at

v = 0 + 9.8 * 9.11

v ≈ 89.36 m/s (rounded to two decimal places)

Therefore, the speed with which the penny strikes the ground is approximately 89.36 m/s.