1. if a 0.05kg coin is hanging from 22m above the ground:
a. If you drop it, how fast will the coin be moving halfway to the ground?
b. How fast will it be moving just before it hits the ground?
If you could point me the right direction with equations that would be great. I have a whole list of equations but I just don't see which ones I am supposed to you. Also, when the problem says "fast" that means velocity right?
fast=velocty
a) KE= change in PE
1/2 m v^2= mg (22/2)
b) KE= change in PE
1/2 m v^2=mg(22)
Did you get 22/2 because it said halfway?
Yes, in this context, "fast" refers to velocity. To solve this problem, you can use the equations of motion, specifically the equations for constant acceleration under gravity. Here are the relevant equations:
1. Equation of motion relating displacement, initial velocity, time, and acceleration:
- s = ut + (1/2)at^2
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
2. Equation of motion relating final velocity, initial velocity, acceleration, and displacement:
- v^2 = u^2 + 2as
where v is the final velocity.
To solve part (a), we need to find the velocity of the coin when it is halfway to the ground. We can use equation (1) since we know the initial displacement, acceleration, and want to find the velocity at a specific time.
a. To find the velocity halfway to the ground, we need to find the time it takes to reach that point. The coin is initially at a height of 22m and will fall a total height of 22m / 2 = 11m. We can use equation (1) to solve for time (t):
- s = ut + (1/2)at^2
- 11 = 0t + (1/2)(9.8)t^2
- 11 = 4.9t^2
- t^2 = 11 / 4.9
- t ≈ √(11 / 4.9)
- t ≈ 1.466 s
Now we can use this time value in equation (1) to find the velocity (v) halfway to the ground:
- v = u + at
- v = 0 + (9.8)(1.466)
- v ≈ 14.3 m/s
So, halfway to the ground, the coin will be moving at approximately 14.3 m/s.
b. To find the velocity just before the coin hits the ground, we can use equation (2) as the final velocity is what we are looking for.
- v^2 = u^2 + 2as
- v^2 = 0^2 + 2(9.8)(22)
- v ≈ √(2(9.8)(22))
- v ≈ √(431.2)
- v ≈ 20.8 m/s
So, just before it hits the ground, the coin will be moving at approximately 20.8 m/s.
Remember to always double-check the calculations and units to ensure accuracy.