"Two rocks are thrown off the edge of a cliff that is 15.0 m above the ground. The first rock is thrown upward, at a velocity of +12.0 m/s. The second is thrown downward, at a velocity of −12.0 m/s. Ignore air resistance. Determine (a) how long it takes the first rock to hit the ground and (b) at what velocity it hits. Determine (c) how long it takes the second rock to hit the ground and (d) at what velocity it hits." I'm frustrated... extremely. I have been trying to figure this out for the past two hours and I know its simple! Edit: new picture, I got A and B right so far, a little lost on C and D Edit: Googling this problem results in a bunch of trap answers setup by different schools/universities. This problem is apparently popular because lots of people (myself included) don't understand what to do.

Ok, so I got A finally after realizing an error a) 3.36 - verified correct. I can't seem to get B though. I've tried 20.93... I know I need to divide by 9.8 (gravity) from the height given of the point when it drops (22.3469) Edit : Got B: -20.93

Sorry, I thought you were referring to C) when you posted that. My bad. C) is actually easier to think about in all positives, since the displacement, initial velocity and acceleration all have the same direction.

Thank you! You really helped me understand a key concept I was missing! If I am not mistaken this leads me to have only 1.53 s for c? Edit: No wait that is wrong... let me try again Edit: I got 2.1, but it said I was incorrect... Trying my math again.

Either use the quadratic formula or use Wolfram|Alpha. This is an EXTREMELY valuable resource for Physics. Quadratic formula if you can't recall: Warning: Spoilers inside! Specifically for quadratic formula: http://www.math.com/students/calculators/source/quadratic.htm

This works on a few problems (using quadratic) In this case I can not, it will result in imaginary numbers. I have to rely on finding T by Vf = Vi + at

Hm? Plug in these values and it comes out fine: a = -4.9 b = -12 c = 15 Remember that you move the -15 to the other side, where it becomes "c".

I am finding "t" the time in seconds though, (see my image near the top left when I solve for t) vfin=vint+at xfin= xinit + vinit*t+ 1/at^2 12*t+1/2*9.8*t^2

Yes, the answers you get by plugging in those three values are the x-intercepts [time (s) values] where the rock hits ground level (Δx = -15 or x = 0). This is our function: -15 = -12t -4.9t² -4.9t² -12t + 15 = 0 or 4.9t² + 12t - 15 = 0 a = 4.9 b = 12 c = -15 or flip the signs on all three, it doesn't change anything.

Yes, my mistake I was looking at the function incorrectly. Ironically right as you posted this I realized my answer 0.911 was the real answer. I disregarded it as impossible for some reason. I felt the value was too small. Argh I feel like an idiot. Not sure why I was getting i in my equation earlier.. Thank you very much!

NP, if you need help on future problems feel free to message me, I probably need to brush up on my Physics I anyway. By the way, just a small shortcut: if you flip only the b value in a quadratic, the x-intercepts will merely change signs. You can test this on math.com, using any a value and c value with opposite signs.

Though you could mess around with negatives and fun equations, they will work with it, why not break it down and make life so much easier? I assume they are not trying the equivalent of a double negative as far as velocities and displacements go. a and b are basic use of equations of motion. c and d are still basic but perhaps best to split into two parts 1 the rise, 2 the fall. The rise then. You need to figure out when the velocity drops to 0 on the upswing, also the height it reaches during that. The fall. We are back at the basics. How long does it take an initially stationary object to fall 15 meters + the earlier rise height under standard Earth gravity, no friction/air resistance. Do remember to add the time taken in the initial rise. Under Earth gravity find how fast an initially stationary object is going from a fall of 15 meters plus initial rise height, no air resistance. No need to add anything here.

jonthedit I wish I'd seen this yesterday, I love physics problems! Were you required to use the Δx = Vot + (½)at² equation? Because the problem is a little simpler if you use V = Vo² + 2ax and V = Vo + at. Here's the approach I would have taken: Warning: Spoilers inside! I solved Rock #2 first because the situation is simpler. You will notice that once Rock #1 falls back down to its original position from where it was thrown upward, it is identical to the situation shown with Rock #2. For Rock #1, it's actually going straight up and straight down, but I just drew a parabolic trajectory so I could show the positions more clearly In most cases, using 10 for gravity (instead of 9.8) is good enough unless your instructor is nit-picky or you're in some intense engineering class. For that last part where I got 3.3 seconds, if you have to show your work, you could use V = Vo + at, but you can actually do it in your head if you get used to doing it the way I explained. So if you shoot a ball straight upward from the ground at 50 m/s, it will take 5 seconds to reach the top (V = 0), and another 5 seconds to hit the ground (10s total air time). It will be traveling at 50 m/s when it lands. If you shoot a ball straight upward at 25 m/s, it will take 2.5s to reach the top (V = 0) and 2.5s to hit the ground (5s total air time), and it will be traveling at 25 m/s when it lands. Since the initial velocity of Rock #1 was 12 m/s up, you know that it takes 1.2s to reach the top (V = 0) and another 1.2s to return to its original position (2.4s total air time), at which point its velocity will be exactly equal and opposite to its initial velocity (now 12 m/s down), assuming no air resistance. It's quick and dirty, but that's how ninjas roll *EDIT* Didn't see FAST's reply before I posted mine, but the difference is that he chose to find the distance on the upswing (use V = Vo² + 2ax), then add it to the final 15m distance to the ground. Multiple ways to approach this problem; you just have to find most efficient method.

I hate you so much, I was just to post the same thing, pretty much >:3 Lets team up and solve people's physics problems and homeworks

LOL! No thanks! I only need help understanding how to do problems occasionally. 'Solving of homework' would lead to no learning.