Physics.....my class is working on this one together with no success....any help would be appreciated.?

Question

A skateboarder moves at v = 4.4 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is h = 0.48 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.

Answer 1

initial speed vo = 4.4 >>motion on inclined plane
Length covered = L = h/sin 48 >> h = 0.48
air-borne or projecttion speed (v) given by
v^2 = vo^2 - 2 (g sin 48) L or v= 3.155 m/s
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projectile shot up at (v initial now), angle of projection (p = 48)
H (max) = v^2 sin^2 48/2g
H (max) = 0.28 m

this Hmax is for a projectile which gets air-borne from h = 0.48 level bcause from there v is taken. Hmax if from h=0.48
from ground = 0.48 + Hmax

Answer 2

because we are solving the maximun height at which the skateboarder reaches, we need to find the vertical velocity of the skateboarder at the end of the track

Vy = sin(48)4.4 = 3.27 m/s


The question says find the maximun height measure from the end of the track (not the ground), so let 0m be the initial position of the skateboarder

Vf^2 = 2ad + Vi^2

when the skateboarder reaches his maximun height, his final vertical velocity is 0m/s

0^2 = 2(-9.8)d + (3.27)^2
-10.6929 = -19.6d
d = 0.5456m


the skateboarder 's maximun height is 0.5456m measure from the end of track.

Answer 3

When she leaves the track, you apply the second law of motion:
Forces = mass*acceleration
the only force against her is gravity: F=mg where m is her total mass and g the acceleration of gravity.

mg =ma so a = g

If her acceleration is g, the vertical component of her speed is:
Vy(t) = Voy - gt where Voy is the vertical component of her initial speed and t is the time.
Voy = Vo*sin48° where Vo = 4.4 m/s
her height would be:
H(t) = Ho +Voyt -1/2gt^2
She reaches her max height when the vertical component of her speed is 0 (it is initially positive as she rises, and negative as she starts falling).
Vy(t) = 0 = Voy - gt, so this happens at the time t when
t = Voy/g
Replace that in the second equation:
H = Ho + Voy*Voy/g -1/2g(Voy/g)^2
H = Ho + 1/2(VoSin48)^2/g
Ho=0.48m, Vo=4.4m/s, g = 9.81 m/s^2
H = 1.02m

Answer 4

One question: does she lose speed due to the rise of 0.48 m? I will assume yes, so her speed on leaving the track is
.5*m*vi^2-m*g*h=.5*m*vf^2
vi^2-2*g*h=vf^2
vf=sqrt(4.4^2-2*9.8*0.48)
vf=3.15468 m/s

Since the ramp has an angle, her vertical component of speed is
sin(48)*vf
=2.3444 m/s

Using conservation of energy again
m*g*h=.5*m*2.3444^2
h=.5*2.3444^2/9.8
h=0.28 m

adding in 0.48 for the height
H=0.76 m

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