2D kinematics problem -- Skateboard ramp jump calculations

• weewooweee
In summary: For V_x, t1 is 0.946 and t2 is 4.26. For V_y, t1 is 0.946 and t2 is 3.58. So the horizontal distance is 4.26 m.
weewooweee
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.

weewooweee said:
for V_x the final is the same as the initial since the equation becomes V_xf = V_xi
Why is that the case? The acceleration is ##a=g\sin\!\theta## down the incline. This has x and y components. You do not have projectile motion while the skateboarder is on the incline.

kuruman said:
Why is that the case? The acceleration is ##a=g\sin\!\theta## down the incline. This has x and y components. You do not have projectile motion here.
using v^2 = u^2 + 2as, since in the horizontal direction acceleration is 0, v^2 = u^2 and they're the same value. Not sure though, haven't taken anything to do with forces so far, only 1D and 2D kinematics.

When the skateboarder moves in a straight line up the incline the velocity vector is along the incline and stays that way until she flies off at the end. This means that ##\dfrac{v_y}{v_x}=\tan(30^{\circ}).## You cannot say that the vertical component decreases but not the horizontal component. If that happened, the skateboarder would plow into the incline. Both components must decrease uniformly to maintain their ratio constant.

MatinSAR and Lnewqban
Gravity is slowing down the skateboarder while on the ramp; therefore, V1 < V0.
Calculating t1 and t2 will help you find the horizontal distance of the flight.

MatinSAR

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