Projectile motion of skateboarder problem

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 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

xf=xo+vot+1/2at2
t=d/v
R=vo2/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 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.

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

why do you take t=d/ v of y direction? if y is vertical direction then it will come to v =0 after a time t so 0= v(y) - gt and acceleration is -g.
one should get H nearly 0.76 m above the point of leaving the ramp.

if v is the velocity of projection from the ramp then v. sin (theta) is the vertical velocity and H= (v sin(theta))^2 /2.g using the relation of final velocity and the height H with initial vertical velocity.

why do you take t=d/ v of y direction? if y is vertical direction then it will come to v =0 after a time t so 0= v(y) - gt and acceleration is -g.
one should get H nearly 0.76 m above the point of leaving the ramp.

if v is the velocity of projection from the ramp then v. sin (theta) is the vertical velocity and H= (v sin(theta))^2 /2.g using the relation of final velocity and the height H with initial vertical velocity.
so then the hight it reaches is 1.31m by using H= (v sin(theta))^2 /2.g assuming that 2.g is meant to be 2 times g? and would it be 1.31above the ramp or would you need to subtract the height of the ramp?

TSny
Homework Helper
Gold Member
I took the top parabola as its own projectile and tried to solve as if there was no ramp or anything,

You can't neglect the ramp. She slows down while going up the ramp.

gneill
Mentor
@Erenjaeger : Can you show us details of your calculation to determine the speed of the skateboarder as she just reaches the top of the ramp (her "launch" speed)? What information are you given about that portion of the skateboarder's trajectory that might help you decide on the what approach to take?