- #1

weewooweee

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- Homework Statement
- A skateboarder starts up 1m high 30 degree ramp at a speed of 5.2m/s. The skateboard wheels roll without friction. At the top, she leaves the ramp and sails through the air. How far from the end of the ramp does the skateboarder touch down?

- Relevant Equations
- v^2 = u^2 + 2as

s= vt - 1/2at^2

s = ut + 1/2at^2

So I tried the following:

Getting the velocities for x and y

V_xi = 5.2cos(30) = 4.5

V_yi = 5.2sin(30) = 2.6

Then I use v^2 = u^2 +2as to get the final velocities before she leaves the ramp:

for V_x the final is the same as the initial since the equation becomes V_xf = V_xi

for V_y the final is the following: v_yf = sqrt(|2.6^2- 2*9.8*1) = 3.58 m/s -> This doesnt make sense to me, shouldnt it slow down?

After this, I use the equation s = ut+ 1/2at^2 to calculate the horizontal distance as:

-1 = 3.58t - 1/2(9.8)t^2

Solving the quadratic gives t=0.946 s

Using s =ut+1/2at^2

s = 4.5(0.946) + 0 = 4.26 m

Which is incorrect, I'm not sure where I went wrong here.

Getting the velocities for x and y

V_xi = 5.2cos(30) = 4.5

V_yi = 5.2sin(30) = 2.6

Then I use v^2 = u^2 +2as to get the final velocities before she leaves the ramp:

for V_x the final is the same as the initial since the equation becomes V_xf = V_xi

for V_y the final is the following: v_yf = sqrt(|2.6^2- 2*9.8*1) = 3.58 m/s -> This doesnt make sense to me, shouldnt it slow down?

After this, I use the equation s = ut+ 1/2at^2 to calculate the horizontal distance as:

-1 = 3.58t - 1/2(9.8)t^2

Solving the quadratic gives t=0.946 s

Using s =ut+1/2at^2

s = 4.5(0.946) + 0 = 4.26 m

Which is incorrect, I'm not sure where I went wrong here.