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.