Free Fall Question [ Need Assistance]

AI Thread Summary
To determine the height from which a pile driver should be released to achieve a speed of 20 m/s upon impact, the equation v² = V(o)² + 2ax is utilized, where v is the final velocity, V(o) is the initial velocity, a is acceleration, and x is the position. The calculation initially leads to a negative position value, indicating an error in unit handling. It's emphasized that acceleration can be defined as negative when considering downward motion, which affects the sign of displacement. Additionally, the use of the equation x = V(o)T + 1/2at² is suggested, but time is not provided in the problem. Proper attention to units is crucial for accurate results.
pharmguy
Messages
8
Reaction score
0

Homework Statement


It is desired to have a pile driver strike a pile with a speed of 20 m/s. Assuming the driver to be in free fall, from how high above the pile should be released?


Homework Equations


Used v^(2) = V(o)^(2) + 2ax
v=final velocity
V(o) =initial velocity
a=acceleration
x=position


The Attempt at a Solution


I set up my problem as (20 m/s)^(2) = (0m/s)^(2) +2(-9.8 m/s^(2))x finding position (x).
Next step looks like this 400 m/s^(2) = -19.6 m/s^(2) (x)
Last step looks like (400 m/s^(2))/(-19.6 m/s^(2)) = x
Answer: -20.4 = x
Answer is negative and units are wrong. I believe all the units would cancel out and the answer would be unitless.

I was also thinking about using the equation x=V(o)T + 1/2at^(2) but problem is I wasn't given a time unless I assume time to be 0.

Please Help, Thank You.
 
Physics news on Phys.org
Your units aren't quite right, v^2 is (m/s)^2 = m^2/s^2
acceleration is m/s^2, not quite the same thing.

You are free to pick any direction for your acceleration and distance, as log as they agree. If you want acceleration down to be negative, that means displacements down are negative.

ps. Extra marks for checking (and noticing) the units ;-)
 
Awesome thanks a bunch! Is there a rating system or something somewhere on this site to rate people's help?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top