- #1

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## Homework Statement

The drawing shows a skateboarder moving at 6.60 m/s along a horizontal section of a track that is slanted upward by

*θ*= 48.0° above the horizontal at its end, which is 0.700 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.

pic:

**https://fbcdn-sphotos-g-a.akamaihd...._=1470375341_7e696449468e4fc14a732118b9cd1b46**

https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-xfp1/v/t34.0-12/13933151_1069671526445955_1581744661_n.png?oh=c77ec7fc191304680811f97a0b612d82&oe=57A35803&__gda__=1470375341_7e696449468e4fc14a732118b9cd1b46

https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-xfp1/v/t34.0-12/13933151_1069671526445955_1581744661_n.png?oh=c77ec7fc191304680811f97a0b612d82&oe=57A35803&__gda__=1470375341_7e696449468e4fc14a732118b9cd1b46

## Homework Equations

x

_{f}=x

_{o}+v

_{o}t+1/2at

^{2}

t=d/v

R=v

_{o}

^{2}/g sin2θ

## The Attempt at a Solution

I took the top parabola as its own projectile and tried to solve as if there was no ramp or anything, just that projectile.

I used that kinematic formula for the Y direction by going y

_{f}=y

_{o}+v

_{oy}t+1/2gt

^{2}

I found the time to put in the equation by going t=d/v

_{y}and d is equal to the range/2 since we want the highest point the v

_{y}came from going 6.6sinθ (θ=48°) to get the vertical component of the velocity and found the distance by plugging the numbers into the range formula and dividing by 2

so R=5.07

^{2}/9.8sin96 and that was 4.372m so divide that by 2 to get 2.186m

so now t=2.186/5.07= 0.43

y

_{f}=0+5.07⋅0.43+1/2⋅9.8⋅0.43

^{2}= 3.086m

Can anyone see anything wrong with this? because it took me a few goes to get to this stage and on my assignment where im submitting this answer we get 5 attempts and ive used 4 attempts to get to this stage haha, any help is good.