1. The problem statement, all variables and given/known data 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 heightH 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 2. Relevant equations xf=xo+vot+1/2at2 t=d/v R=vo2/g sin2θ 3. 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 yf=yo+voyt+1/2gt2 I found the time to put in the equation by going t=d/vy and d is equal to the range/2 since we want the highest point the vy 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=5.072/9.8sin96 and that was 4.372m so divide that by 2 to get 2.186m so now t=2.186/5.07= 0.43 yf=0+5.07⋅0.43+1/2⋅9.8⋅0.432 = 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.