Skateboard Ramp Kinematics Practice Problem

[jehan4141]

This is an even problem with no solution provided. My quiz is Monday. Could somebody check this practice problem for me?

A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58° above the horizontal. The end of the ramp is 1.2 m above the ground. Let the x-axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp.


(a) How high above the ground is the highest point that the skateboarder reaches?
(b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

V_o = 6.6 m/s
V_oy = V_osin58 = 6.6sin58
V_ox = V_ocos58 = 6.6cos58

V_y = 0 m/s at the highest point

V_y² = V_oy² + 2ay
y = ( V_y² - V_oy² ) / 2a
y = [0 - (6.6sin58)2] / (2*-9.8)
y = 1.598353244 m

y + 1.2 = 2.798 m

V_Fy = 0 m/s
V_oy = 6.6sin58

V_Fy = V_oy + at
t = ( V_Fy - V_oy ) / a
t = ( 0 - 6.6sin58 ) / -9.8
t = 0.5711344321 s

V_x = x/t
x = Vx_xt = (6.6cos58)(0.5711344321)
x = 1.997 m

 

[Delphi51]

They all agree with my answers.

 

[jehan4141]

Thank you!