Alright, i did this and got a value of 3.43 for Voy. after plugging the values into the equation my value for y was 0.6 m.
Thank you SO much for your help, this has been really helpful and has cleared up a lot of questions i had about Kinetic Energy and such. Cheers!
okay bear with me here lol. so with that value i can use kinematics to solve for the rest right?
Vo = 4.617 m/s
ay = -9.8 m/s
Vf = 0 m/s
then solve for y?
Oh okay, yeah stupid mistake. Alright now I'm getting a value of 4.6 m/s. So now I'm supposed to move on to the next section of her jump using this as the initial velocity?
so would the velocity at the end be:
1/2(m)(Vo)^2 + mgho = 1/2(m)(Vf)^2 + mghf
cancelling the m's, and isolating for Vf:
Vf = (2 (1/2(Vo)^2 - ghf))^1/2
if so, i calculated vf (at the end of the ramp) to be 5.02 m/s
Thats one of my main questions... would that value be 5.4/cos48? so 8.07 m/s?
So with that in mind, when dealing with a problem involving KE, does the orientation of the velocity matter?
Homework Statement
The drawing shows a skateboarder moving at 5.4 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is 0.40 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion...