physics help

Elin · May 2011

I'm attempting to do my physics homework and it's KICKING MY ASS. I thought I knew what was going on, but I was wrong, wrong, wrong.

A force F1 of magnitude 6.70 units acts on an object at the origin in a direction θ = 37.0° above the positive x-axis. A second force F2 of magnitude 5.00 units acts on the object in the direction of the positive y-axis. Find graphically the magnitude and direction of the resultant force F1 + F2.

This is a sample problem like one of my homework problems, it's using different numbers.

I thought that you found all the forces working in the Fy, and all in the Fx.
Then added those for the magnitude.

And then took the arctan of Fy/Fx for the direction, but ... either I'm calculating it wrong or that's not how you solve it.

----

A rock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal range R

At what angle θ is the rock thrown?

In terms of its original range R, what is the range Rmax the rock can attain if it is launched at the same speed but at the optimal angle for maximum range?

This one ... I haven't even got a clue how to start. How do I find an angle when there's no real info given.

I mean, if you have the h (opposite) and the R (adjacent) the answer to part one would be arctan(H/R), right? The second part .. I have no clue.

The homework is online using webassign, so there's no answer in the back of the book I can even reference. Can someone point me in the right direction?

Dunadan019 · May 2011

magintude is (X^2 +Y^2)^.5

direction is arctangent of (Y magnitude/X magnitude)

tarnok · May 2011

Ugh. This would be much easier to explain if I could draw pictures for you. The way you're trying to do it doesn't work though. You don't just add the components in the x and y directions for the magnitude. When adding vectors you set the tail of one vector at the head of the next, then draw a new vector from the tail of the first to the head of the second.

Here. I think you want the explanation of adding vectors in polar form on that page.

Dunadan019 · May 2011

for the second one, you are supposed to have the equations for simple trajectories which are:

Range = ((Vi^2)*sin(2Theta))/g
Max height = ((Vi^2)*sin^2(Theta))/2g

tarnok · May 2011

I think for the second one you're looking for the angle of the tangent to the parabola at the first x intercept. Or use those equations I guess.

Demerdar · May 2011

Isn't the maximum range just taking the"range" equation, taking its derivative with respect to theta and set it equal to zero? This should give you the maximum of the range function in terms of theta (and therefore is your "optimal angle"). When they say, "in terms of" it means most likely as a ratio between R_max and the original range (R), such that R_max=~1.3R (or whatever).

And don't worry about having trouble with it, the connection between physics and calculus is a little bit hard to grasp at first.

Elin · May 2011

Demerdar wrote: »

Isn't the maximum range just taking the"range" equation, taking its derivative with respect to theta and set it equal to zero? This should give you the maximum of the range function in terms of theta (and therefore is your "optimal angle"). When they say, "in terms of" it means most likely as a ratio between R_max and the original range (R), such that R_max=~1.3R (or whatever).

And don't worry about having trouble with it, the connection between physics and calculus is a little bit hard to grasp at first.

It would be easier if it were calc, this is algebra based. I should have specified that.

The prof told me by email that I should: For #5, if you look in the book, you can find the range equation. That will tell you the range R (same thing as delta-x). Now you need to find an equation to describe the vertical height. Set the maximum height equal to R and solve for theta.

This still makes no damn sense. I need a tutor.

Dunadan019 · May 2011

Elin wrote: »

Demerdar wrote: »

Isn't the maximum range just taking the"range" equation, taking its derivative with respect to theta and set it equal to zero? This should give you the maximum of the range function in terms of theta (and therefore is your "optimal angle"). When they say, "in terms of" it means most likely as a ratio between R_max and the original range (R), such that R_max=~1.3R (or whatever).

And don't worry about having trouble with it, the connection between physics and calculus is a little bit hard to grasp at first.

It would be easier if it were calc, this is algebra based. I should have specified that.

The prof told me by email that I should: For #5, if you look in the book, you can find the range equation. That will tell you the range R (same thing as delta-x). Now you need to find an equation to describe the vertical height. Set the maximum height equal to R and solve for theta.

This still makes no damn sense. I need a tutor.

Range = ((Vi^2)*sin(2Theta))/g
height = ((Vi^2)*sin^2(Theta))/2g

set those two equal and solve for theta

part 2: the max range occurs at 45 degrees. you can figure this out by simply looking at the above equation and figuring out what value of theta makes the sin(2theta) = to 1 (since that's the maximum of a sin function).

I can't make it more explicit without solving it for you, I suggest you do get a tutor or just some tutoring if this stuff doesn't make sense.

Penny Arcade

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