Recent content by iceman99

Thanks for the help everyone. I really appreciate it.

Homework Statement if the angle of the slope is increased, the change of gravitational potential energy between the two heights: (Select all that apply) (It may be more that one option) increases depends on the path followed. decreases. depends only on the difference between the two...

Ok, thank you very much, I must have written the answer down incorrectly.

In my g, I forgot the square so it should be (2x^2-3x)^-6 and consequently would make g’=-6(2x^2-3x)^-7(4x-3) correct?

No, I do not expect people to be waiting around for me to post a problem; I was merely annoyed that no one had replied in my five hours. Rather than the quotient rule is all I meant by my phrase.

the -14x^2 in my final answer should be -24, but yeah that does not change much.

Find the derivative of y=(3x+4)^5/(2x^2-3x)^6 Some how the answer is simplified to 3x(6x^2-59x+24)/(2x-3)^7, I don’t know if I just copied it down wrong or I’m doing it incorrectly. Some assistance would be nice. I changed the equation to y=(3x+4)^5*(2x-3x)^-6 and used the product...

I can't really believe I was that dumb. Sorry, thanks a lot for the help

Oh I got it

R=(Vo^2sin(2theta))/g i found theta to be 14.607degrees...

OK, but how could I find the max height of the larger angle? Wouldn't the velocity need to be changed or am I wrong about that too?

Um...I had no idea about that. How could I find the larger angle?

Yeah I think that from the wording it means that it was launched from the ground at 30m/s which the maximum height was 2.96m. But because the question doesn't indicate that it wasnt launched from a building, it could be where the 43m comes from.

Yeah that's the whole problem

Just tried another thing. I set out to find the angle of the projectile when it landed setting my right triangle to: Vx=30m/s Vy=14.7m/s found by Vy=9.8m/s^2 x 1.5s Vy-14.7 Found the angle to be 26.10degrees Making a new triangle with 26.10 as my theta and 45m as its adjacent length...