i try i tryOriginally Posted by ProjectWidow
i try i tryOriginally Posted by ProjectWidow
Val for President
shit I still have like 5 more problems left, anyone?
Shoot again, but see if theres formulas, i cant remember them for shit lol.
A 2.3 kg rock initially at rest loses 403 J of potential energy while falling to the ground. Assume that air resistance is negligible.Originally Posted by ironchef
(a) Calculate the kinetic energy that the rock gains while falling.
___J
(b) What is the rock's speed just before it strikes the ground?
___m/s
The main formula is PE(before) + KE (before) = PE (after) + KE (after)
where PE = mgh
and KE = 1/2mv^2
and another one that should be easy
A cannonball is fired at 50 m/s at an angle of 30 degrees. How high does it go? (Hint: determine the vertical velocity, and use conservation principles.)
___m
I just cant seem to figure it out. I tried doing trig functions but it doesnt work out
					
						
					
				
			
			 
			
				i dont remember exactly, but doesnt Potential energy just turn into kinetic energy once it starts moving, therefore a. would be 403 J. Its been a little while, so thats probably wrong, but who knows.
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oh shit!! It was so easy, it was hard. Thanks broOriginally Posted by D16Civic
I would give you rep points but it says I have to spread some love first
					
						
					
				
			
			 
			
				is b. 18.72 m/s
PE=KE=403=.5(2.3kg)v^2
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figured out how high the cannon goes. was using cosine function instead of sine
Was it 127?Originally Posted by ProjectWidow
you would think, right? But I just checked and thats wrongOriginally Posted by D16Civic
no, it went 31.887Originally Posted by ironchef
since you find the sine function which is 50sin30
then you use 1/2w^2=gh (KE=PE)
woops, you are right man! My bad. i put in 18.27 instead of 18.72.Originally Posted by D16Civic
Thanks guys!!
Wow my physics skills have gone to crap, and to think i have to take physics sometimes in the next 3 semesters.
hey its alright. Look at me! Im in physics and cant seem to get a few easy problems done. It just sucks not being able to pay attention at the end of the dayOriginally Posted by ironchef
Yea I hear ya, I was barely keeping my head up in history today, its not even that its so boring, but its just towards the end of the day and im just like let me leave haha.Originally Posted by ProjectWidow
exactly the way I am, lolOriginally Posted by ironchef
alright last one for the night
Justin throws a 12.0 g ball straight down from a height of 2.0 m. The ball strikes the floor at a speed of 7.5 m/s. What was the initial speed of the ball? Assume that air resistance is negligible.
____m/s
Im going to go with either 9.76 or 0 being that it might be a trick questionOriginally Posted by ProjectWidow
					
						
					
				
			
			 
			
				im saying its 4.12 m/s
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you are the fuckin man!! Whenever I can give you more rep points, you got em!Originally Posted by D16Civic
thanks guys, you both were a big help!!
					
						
					
				
			
			 
			
				no problem, glad i could help
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How did you solve that one?
					
						
					
				
			
			 
			
				i just used the formula v(final)^2=v(initial)^2 + 2*a*distance where you are trying to find v(initial). The mass isnt needed since objects will fall at the same rate no matter what their mass is. So then its just plugging in the numbers.
Square root of [(7.5)^2 - 2(9.81)(2)]
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Whoopsie i think i added the 2*9.8*2 instead of subtracting,