Answer:
We need an equation for the horizontal motion and another for the vertical motion.
First, let's look at the vertical motion.
Once the skateboarder is on the air, the only force acting on him/her will be the gravitational force, then the vertical acceleration is the gravitational acceleration.
Ay(t) = -32ft/s^2
Where the negative sign is because this acceleration pulls down the skateboarder.
To get the vertical velocity, we need to integrate over time, so we get:
Vy(t) = (-32ft/s^2)*t + v0
Where v0 is the vertical velocity.
To know the vertical velocity, we use the fact that:
When the skateboarder jumped, the velocity was 52ft/s, and the angle was 30°.
Then the initial velocity in the y-axis (the vertical one) as:
Vy = 52ft/s*sin(30°)
and the initial velocity on the x-axis is:
Vx = 52ft/s*cos(30°)
Replacing that in the vertical velocity equation:
Vy(t) = (-32ft/s^2)*t + 52ft/s*sin(30°)
We need to integrate again to get the position equation:
Py(t) = (1/2)*(-32ft/s^2)*t^2 + 52ft/s*sin(30°)*t + p0
Where p0 is the initial vertical position, this would be the height of the ramp, which is 5ft
Then the equation of the vertical motion is:
Py(t) = (1/2)*(-32ft/s^2)*t^2 + 52ft/s*sin(30°)*t + 5ft
For the horizontal motion, we don't have any force, then the acceleration is:
Ax(t) = 0
The velocity is:
Vx(t) = 52ft/s*cos(30°)
And the position equation is:
Px(t) = 52ft/s*cos(30°)*t + p0
Where p0 is the initial horizontal position, we can assume that is zero, then:
Px(t) = 52ft/s*cos(30°)*t
Then the two equations that represent the skater jump are:
Py(t) = (1/2)*(-32ft/s^2)*t^2 + 52ft/s*sin(30°)*t + 5ft
Px(t) = 52ft/s*cos(30°)*t
