Recent content by Eggphys

Just bumping back to the top.

So i decided to come at this from another angle. First i solved for Fg, Fg=1.57*9.8 = 15.39 Then to solve for Fm, Fm=15.39cos 13.2 = 14.98 But the correct answer is 67.38. I am not sure where to go from here to get the correct answer. Any help would be appreciated. Thanks:smile:

Homework Statement There are essentially three forces involved in flying: (i) ``lift'' Fl generated by the difference in air pressure between the top and bottom parts of a wing (or wing-like object), (ii) ``drag'' Fd due to air resistance, and (iii) ``gravity'' Fg. Note that ``lift'' is...

After going over my work abit more I see I have no clue what I am doing:frown: Im not sure how to implement the angle into the equation or what a is in the x direction(Isnt it 0?) Would Fy just be Fy=1.56*9.8? Any help on where to start would be great. Thanks:smile:

Homework Statement The main muscle responsible for raising an arm is the deltoid. The deltoid muscle connects at the upper end of the shoulder, extends over the upper arm bone (humerus), and attaches near the elbow. Effectively there are three forces involved in raising the arm: (i) the force...

Sweet, I appreciate the help mate.

Oh, a in the x direction is 0 right? So that make's the equation 6.06=2.6Vo which gives Initial Velocity of 2.33. Which is correct right?. Haha, I can't believe it was that simply of a mistake. Thanks mate:smile:

1/2at^2 so 4.9*2.6^2

I have another projectile motion problem and instead of making another thread ill just update this one. 1. Homework Statement At Acapulco, professional divers jump from a 33.1 m high cliff into the sea. At the base of the cliff, a rocky ledge sticks out for a horizontal distance of 6.06...

Thanks a lot mate, I see where I went wrong with the algebra. I really appreciate it:smile:

Thanks for the quick response and cheers for the welcome:smile:. I'm not quite sure how to do each steps. For finding the time to take to fall I used X=Xo+Vot+(1/2)at^2 then adding in the values 0=1.5+4.9t^2 and getting the answer 0.25 then using the same equation for distance traveled...

Homework Statement An archer shoots an arrow over level ground. The arrow leaves the bow at a height of 1.50 m with an initial velocity of 79.5 m/s in a horizontal direction. At what horizontal distance (in m) does this arrow strike the ground? The answer is 44m but I don't know how to get to...