A penny is dropped from the observation deck of the Empire State building (370 m above ground). With what velocity does it strike the ground? Ignore air resistance.
answer is in____ m/s.
-85.158
vf^2=vi^2+2g*distance
please elaborate. this equation is useless. answer the question
A penny is dropped from the observation deck of the Empire State building (370 m above ground). With what velocity does it strike the ground? Ignore air resistance
To determine the velocity at which the penny strikes the ground, we can use the principles of motion and gravity. Let's break it down step by step:
1. Determine the initial velocity: Since the penny is dropped from rest, the initial velocity (u) is zero m/s.
2. Determine the final velocity: The final velocity (v) can be found using the equation:
v = u + gt
Where:
v = final velocity (which we want to find)
u = initial velocity (which is 0 m/s)
g = acceleration due to gravity (which is approximately 9.8 m/s^2)
t = time taken for the penny to fall
3. Calculate the time taken to fall: We can calculate the time (t) it takes for the penny to fall using the following equation derived from the kinematic equation:
s = ut + (1/2)gt^2
Where:
s = distance fallen (which is 370 m)
u = initial velocity (0 m/s)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken (which we want to find)
Rearranging this equation, we get:
370 = (1/2)(9.8)t^2
Simplifying it further, we have:
t^2 = (2 * 370) / 9.8
t^2 = 74.898
t ≈ √74.898
t ≈ 8.66 s (approximately)
4. Plug the values into the equation for final velocity:
v = u + gt
v = 0 + (9.8 * 8.66)
v ≈ 84.888 m/s
Hence, the velocity at which the penny strikes the ground, ignoring air resistance, is approximately 84.888 m/s.