The drawing shows a skateboarder moving at 7.06 m/s along a horizontal section o

Question

The drawing shows a skateboarder moving at 7.06 m/s along a horizontal section of a track that is slanted upward by 38.9

The drawing shows a skateboarder moving at 7.06 m/s along a horizontal section of a track that is slanted upward by 38.9 degree above the horizontal at its end, which is 0.703 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.

Explanation / Answer

by coservation of energy we get the velocty of the girl at the height when it leaves the wall

so 1/2mv^2 +0 = 1/2mV1^2 + mgh

= 24.9 J = 1/2v1^2 +6.8

v1=6m/s

therefore now for the max. vertical height reached we know that it will depend upon the the vertcal component of velocity at the point when it leaves the wall

so at max height the velocity = 0

so v= u +at

0=6sin 38.9 -9.8*t

t=0.38 s

so now the height traversed = s= ut -.5at^2

=s= 6sin 38.9 *t -5*9.8*t^2

s=.72 m above the hieght so the toatal height above the ground = .72 +.703 = 1.423m

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