Posted by **Mary** on Wednesday, February 28, 2007 at 7:39pm.

The drawing shows a skateboarder moving at v = 5 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is h = 0.52 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to which she rises above the end of the track.

bobpursley I really appreciate all of your help so far, but I am still confused about this question. This is what I've done so far but it is wrong:

mgh= 1/2m x 25 x (sine 48)^2

I solved for h

h = 0.5m x 25 x 0.552/ mg

the m cancels out leaving

h = 0.5 x 25 x 0.552/ 9.81

h = 0.7037

and to find H

H = h - 0.52

H = 0.7037 - 0.52

H = 0.1837

Please help!!!!!!!

Thanks, Mary

It looks right to me. The only question is significant figures. the velocity is given as one significant figure, and h is given as two. Clearly, your answer to four digits is wrong. Technically, the answer ought to be to one digit(.2meters), however, I am not certain your instructor is a nit picker on that. Your work is on target.

Ok. First you take the derivative of the selected quadry, half-end both secular connotations PL-ing the quantum Chancery's over your answer. Simple.

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