Recent content by sundrops

here's what i did for this problem: (Vy)^2 = (Vyo)^2 - 2g(y - yo) 0 = (Vyo)^2 - 2(9.8m/s^2)(3.5m-0m) Vyo = 8.35m/s my answer was incorrect however, I was wondering where I went wrong...

it's wrong. So obviously my method of getting the answer was flawed aswell. Is there another way to solve this problem?

i guess that makes sense - see I thought of it as the plane would drop the supplies directly over the sled but the supplies don't drop straight down but dop at an angle so i drew a triangle. but i guess plane and sled going the same way makes more sense;)

A swimmier heads across a river, swimming at 1.20m/s relative to still water. It arrives at a point 67.0 m downstream from the oint directly across the river, which is 100.m wide. What is the speed of the river current? Here's what I did: tan^-1(67/100) theta = 33.82 degrees Vx =...

alright - I'll try that

A cargo plane is dropping supplies to a group of people dogsledding acoss Antarctica. The plance drops the package when it is directly above the group of people at an altitude of 1950m. The dogsled travels at 5.68m/s and reaches the supplies in 6.70mins. What was the speed of the cargo plane...

YAY! It's right! Thanks SOOOO much vincent! you are a lifesaver - or at least a HUGE stress reliever! lol

okay - that makes sense. I keep getting 8.65m/s

g wpuld be a negative number right? (-9.8m/s^2 as opposed to 9.8m/s^2)

oh gosh - okay - I'm sure I can - thanks for the help - I'll post my answer if I ever get around the 5 range! lol :redface:

well that didn't work out... here's how I manipulated the equation: Vo = Square root of { tan(38)x - ((-g)x^2 / 2ycos(38)^2) }

Vo = \sqrt{(tan(38)(7.0m) - \frac{(-9.8m/s^2) (7.0m)^2}{2 (0.3m) cos^2(38)}}

ok well at least I know the approximate range of the answer right? hehe

that answer didn't work either what am I doing wrong? ~ I only have 2 tries left :S

does that sound more feasible?