Recent content by scroobnab

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?

also, sorry i don't have the picture attached, I am new to the site and not even sure how to do that..

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...