Recent content by solarwind

Hi all. Let's say I have a set of data as follows (the mass of a sample of some chemical measured several times): 23.132 g 24.532 g 21.532 g 22.853 g 23.193 g (I just made that data up, but imagine that a analytical scale put out those numbers, exactly as shown, on its display.)...

And where exactly did you modify it so it takes into account that gravity is pointing down? Edit, nevermind I see it. Thanks so much!

No, but I know your equation is right, I simulated it.

Thank you! It works now. Why is it not negative? And thanks again for all your help! It's clear now!

According to your equation, I did this: Solve[0 == -(14 + 0.08 (-9.81)) + 10 u - 0.08 (-9.81) u^2, u] It spits out: {{u -> -13.9493}, {u -> 1.20716}} So far, I have 3 different sets of answers (6 answers in total) which all seem to work out. There should be only 2 solutions. Which one is right?

Well sorrrrryyyyyy! Can you explain how you got from the top two equations to the quadratic equations? I tried it but I'm getting imaginary numbers.

I put in: Solve[25^2 == (10/t)^2 + ((14 + 4.9 t^2)/(t))^2, t] It gave me: {{t -> -4.43739}, {t -> -0.791265}, {t -> 0.791265}, {t -> 4.43739}} These two are the right answers for t: {t -> 0.791265}, {t -> 4.43739} All I want to know is how to get to this point on paper.

Ok now I have all the equations I need: V[x]^2 + V[y]^2 = 25^2 = 625 14 = V[y] \Delta t - 4.9 t^2 10 = V[x] \Delta t I know how to do simultaneous equations. I have to solve for V[x] and V[y] from the second and third equations and plug that into the first one to get time. Then I can...

[/B] That's exactly what I did but I don't know where to go from there.

Haha! I know how that is... The answer is NO. This is because x could have a positive displacement and could also have a negative acceleration (slowing down) at the same time. Example, you roll a ball on the floor. It moves away from you (positive displacement) but is also slowing down...

For your first question, what is x? Is it displacement, velocity or what?

No, depends on reference. You can say gravity has a negative acceleration in the y prime, but a falling object is accelerating. The correct term is negative acceleration, not decelerating. Note: I could be wrong.

Projectile Motion - Finding The Angle Homework Statement You are standing 10 m away from a basketball net that is 14 m above the ground. Assume delta d[y] is 14 m, do not account for your height. You launch the ball with a velocity of 25 m/s and it goes in the hoop. At what angle(s)...

Ok, I did this: a) Eg = mgh Ek = mv^2, where v I put 260 m/s Et1 = Eg + Ek b) Same thing as above but with the second values. Elost = Et1 - Et2. The answer is supposed tob e 5.5 x 10^9 Joules but i get 4.98 x 10^9 Joules. What did I do wrong? Did I go about this the right way?

Thanks dude!