Recent content by ssjcory

Homework Statement A golfer hits a shot to a green that is elevated 3.00 m above the point where the ball is struck. The ball leaves the club at a speed of 14.0 m/s at an angle of 40 degrees above the horizontal. It rises to it's maximum height and then falls down to the green. Ignoring air...

So I was just a tad bit off for the integration (meant to put the negative in front of the v^-1) So once we have the basic form -3T = -v^-1 Which I think I was able to get to. You have to add the initial velocity since that is where the equation starts? I'm guessing that's something...

Wouldn't the integration of -3 be -3X(or -3T in this circumstance?) And on the right side v^-2 should become v^(n+1)/n+1 ... Yielding (v^-1)/1 ... or v^-1 ? So -3T = v^-1? And plug in the final and then initial velocity? -3T = 1.5^-1 === -.222 -3T = .75^-1 === -.444 This is very...

I am definitely missing something big here. I see that \displaystyle a=\frac{dv}{dt}\,. and I know \displaystyle (-3.00)v^2=\frac{dv}{dt}\,. because we are given \displaystyle a=(-3.00)v^2 As you said divide each side by v^2 (I guess so that we have all the variables on the right. So...

I did differentiation and integration in calc 1 but I'm really not sure how to apply it here.

Homework Statement The problem is labeled Variable Acceleration. The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared and is given (in SI units) by the expression Ax = -3.00 Vx^2. If the marble enters this fluid with a speed of 1.50 m/s, how long...

Thank you very much I finally figured it out. So in the equation when I always have gravity as negative it works. I think I had tried it before that way but in writing a step of the equation I had left numbers off or changed signs... It would seem I need to be extremely careful with making...

Bump: Can anyone tell me if the methodology used to check my answer is correct? It seems like it would make sense but I don't like that my answer doesn't agree with the professor's.

OK I lied ... I stayed up and tried it again... I have a value that is close for distance... 45.04308602 meters(all values kept in calculator memory so not rounded)... His answer is 41m I wonder if it could just be a difference in rounding or something... I'm not sure :( When I plug my time...

Thank you I was wondering if that might have been my issue. I flipped the signs on acceleration but I guess all of them have to go in the opposite directions. I will retry and post my work here. Thank you! EDIT: Tried changing the sign on one of the velocities... came out with wrong answers...

Homework Statement An object is thrown downward with an initial speed of 10 m/s from a height of 60m above the ground. At the same instant, a second object is propelled vertically upward from ground level with an initial speed of 40 m/s. At what height above the ground will the two objects...