
#1
Feb2810, 05:32 PM

P: 3

Suppose a 136 kg manhole cover is launched 50 feet in the air. What is the speed of the manhole cover when it hits the ground on its descent, in m/s and mph?
I converted 50 feet to 15.24 meters, but I do not know what equation to use here. I understand that acceleration due to gravity is 9.8m/s^2. I just don't know how to find the speed when I am only given acceleration due to gravity and distance/height. 



#2
Feb2810, 05:57 PM

P: 45

You can use distance and gravity to find out how long it takes the manhole cover to hit the ground, and then use another equation to turn time and distance into velocity.




#3
Feb2810, 06:08 PM

P: 3

I'd pretty much already determined that, but you've still not gotten me any closer to solving the problem. What equation do you suggest I use for finding time, if that's even what I need to do in this problem?




#4
Feb2810, 06:20 PM

P: 45

Finding speed of an object when it hits the ground, after free fall.
Actually, I was wrong you don't need time at all:
[tex]v_f^2 = v_i^2 + 2ad[/tex] and, of course, vi will be 0. 



#5
Feb2810, 06:48 PM

P: 3

Actually, you were right; I did need time. It's the equation t = Square root of (2y/g). I just plugged in 15.24 meters for y, the height, to solve for t. After solving for t, I used the equation, vf = initial velocity + (a)(t) to solve for final velocity. Thanks for the help.



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