Recent content by EaGlE

i see how you got 45 degrees now, i been drawing the wrong picture ! yea I am more comfortable with using trigonometry, makes more sense now. A.) Find the magnitude of the velocity of the canoe relative to the river. i got -.25301(i)^2 - .29698(j)^2 = C^2 (not sure if both should be...

can someone tell me the easiest way to solve projectile motion problems? like how to read and start it? it is extremely hard for me, i just don't get it.

\mathbf{v}_\mathrm{CR}& = \frac{v_\mathrm{CE}}{\sqrt{2}}\mathbf{i}-\frac{v_\mathrm{CE}}{\sqrt{2}}\mathbf{j}- v_\mathrm{RE}\mathbf{i}. \end{align} ^hmmm, why do i need to sqrt it by 2? and why is vector i and j the same? Vc/r = Vc/e - Vr/e vc/r = 0.420m/s - 0.55m/s im sorry, but i am...

how did you get 45 degrees?

A canoe has a velocity of 0.420 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.550 m/s east relative to the earth. A.) Find the magnitude of the velocity of the canoe relative to the river. B.) Find the direction of the velocity of the canoe relative to...

thank you, works perfectly

Two piers, A and B , are located on a river: B is 1500 m downstream from A . Two friends must make round trips from pier A to pier B and return. One rows a boat at a constant speed of 4.00 km/h relative to the water; the other walks on the shore at a constant speed of 4.00 km/h. The velocity...

i don't remeber us doing this in class at all, so i read the chapter relating to this subject many times, and it's confusing for me. ok i know that a = 9.8m/s^2, initial velocity = 0, initial height = 2r so... x(t) = 2r + 0 + 1/2(-9.8)t^2 2r = -4.9t^2 sorry but i have no clue on...

A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free fall, the ball has a centripetal acceleration of magnitude 2g. 1.)How far from the bottom of the chute does the ball land? Your answer for the distance the ball travels from...

anyway, i solved for t and got 0,6.4489s then plugged t in the position formula agian x(t) = x(0) + v(0)t + 1/2at^2 and got 16.30251507 but the problem only wants the part above the buliding, so i minus 16.3 from it and got .0025 is that correct? (seems incorrect)

A man stands on the roof of a building of height 16.3m and throws a rock with a velocity of magnitude 31.6m/s at an angle of 28.1 degrees above the horizontal. You can ignore air resistance A.) Calculate the maximum height above the roof reached by the rock. Take free fall acceleration to...

Firemen are shooting a stream of water at a burning building using a high-pressure hose that shoots out the water with a speed of 25.0 m/s as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation \alphaof the hose...

(i know for sure the answer for A.) is 3secs) actually, the motocyle has a two times the velocity of the car. so would i divide 18/2 = 9. units don't matter at this online submission, but it's good to know. 36 was the wrong answer. i only get 2 tries left on this question, so I am asking...

your right, i never knew that velocity was speed. lol. it makes things wayyy easier. ok i found A.) it's 3 secs ok working on B now. i can't seen to find the initial speed of the car. using the formula. v(t) = v(0) + at v(t) = 0 + (6m/s^2)(3) so... v(t) = 18m/s^2 is the speed of the...

ok let's work on problem A. first.i read your post many many times and read the question many times and tried to answer it by myself, but i just don't get it. i really don't know what to do, I am sooo confused. what extactly is velocity? how do you spot it from a question? "Speed is distance...