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ok, i understand that it is a velocity, but where is the square root of gd/2 coming from? anyone?

So from the equation that you posted, they are deriving vi sintheta is greater than or equal to sqart (gd/2)?

ok, i understand that it is a velocity, but where is it coming from? why is is the square root of gd/2?

Homework Statement See picture please Homework Equations The Attempt at a Solution Here is what I understand: vi sin(theta) = initial velocity height on velocity/time graph What I don't understand: sqrt(gd/2) is this gravity times initial distance height of target? Where is the second...

Thanks for the response. Now I have one more question about that. In the attachment, the book talks about (1-cos(theta))=2sin^2(theta/2) gives the magnitude of R as R=2Asin(theta/2). I don't see how they jump from the first equation to the second. Can anyone explain?

Homework Statement what is the resultant vector of an isosceles triangle? Homework Equations R^2=a^2+b^2-4abcos(theta) The Attempt at a Solution The books answer R=2acos(theta/2) Using the formula above, and knowing that a=b in a isosceles triangle I am getting...

I am trying to teach myself physics. I have been going through the book "physics for scientists and engineers" by serway and jewett and doing all the problems in the back of the book and comparing my answers with the solution manual. I also downloaded all the physics lecture from MIT and Yale...

Homework Statement Liz rushes down onto a subway platform to find her train already departing. She stops and watches the cars go by. Each car is 8.60 m long. The first moves past her in 1.50 s and the second in 1.10 s. Find the constant acceleration of the train. Homework Equations...

Homework Statement The integral from v=vi to v of v^-2dv = -3 integral from t=0 to t of dt Homework Equations The Attempt at a Solution I am getting -1/v + C=-3t and the book is getting -1/v+1/vi=-3t. Not quite sure where they are getting 1/vi as I am getting C.

Ok, that made complete sense. I am a bit rusty on my math and forgot when you factor "t" out you get two solutions. But i am having trouble visually imagining this. Say you visually imagine x position graph. If there is no change in x horizontally (when x(initial)=x(final)), you would have a...

ok, so what I said is correct for horizontal movement but not necessarily true for vertical moment, correct? I still do not see how the book is getting the time formula from observing that equation.

So my physics book says "from x(final)=x(initial)+V(initial)t+1/2at^2 observe that when x(final)=x(initial), the time is given by t=-2V(initial)/a" Now if change in x is equal to 0, wouldn't your velocity v(initial)t from above along with your acceleration also be zero? How is the book getting...

So h=1/2gt^2. I am having some trouble understanding this equation. Here are my questions: 1. I know you get two completely different answers if you plug in a set value, but why can't you measure height of a free-falling object by multiplying the number of seconds^2 times 9.8m/s^2? 2. Is...