Recent content by Osbourne_Cox

Ok, thank you

y component=50133.87 x component=-24833.87

Anyone?

[b]1. Hi. This is just a detail that I am hung up on, and can't remember from high school. For my final angle calculation, theta=tan^-1(fy/fx), I get a negative angle, -63.65. I know my numbers are correct as part a of my problem was correct, I just can't remember how to make this angle positive...

Instead of giving the answer, it will be better to show the actual calculations by substituting the values. You can rewrite the expression as tan58 = Vo/ax*t - ay/ax. Now solve for t. This formula gave me the right answer.

okay, so using that I get t=0.426s

Yes, and as a result I have spent a week being confused by this problem...

so i get -1.74=t do we just take the absolute value for the time? or have I made another mistake

vi-at=tan 58 (at) -(-9.8)t=tan58(2.1)t-5.6 9.8t=3.36t-5.6 6.44t=-5.6 t=0.87 but time can't be negative..?

Oh, it is very possible, ha ha. I am not good at this stuff at all.

Initially (at time t = 0) a particle is moving vertically at 5.6 m/s and and horizontally at 0 m/s. The particle accelerates horizontally at 2.1 m/s2 . The acceleration of gravity is 9.8 m/s2 . At what time will the particle be traveling at 58◦ with respect to the horizontal?

Degrees, and my calculator was set on degrees.

Okay, so what do I do with those equations? My prof said it should be as easy as Tan 58=(vi-ayt)/axt and whether you need to break that down more or not, I don't know.

d=1/2*ay*t^2 ?

How do we have negative time though? (its the math to a physics problem)