Askateboarder shoots off a ramp with a velocity of 7.3 m/s, directed at an angle of 60° above the horizontal. the end of the ramp is 1.5 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?

y maximum   3.54 m, value X  2.35 m

Explanation:

We have a projectile launch problem, let's calculate the maximum height of the projectile, where the vertical speed must be zero

        Vyf² = Vyo² - 2g (Y-Yo)

Where Yo is the initial height of the ramp 1.5 m

        0 = Vyo² -2g (Y-Yo)

        Y-Yo = Voy² / 2g

       Y = Yo + Voy² / 2g

Let's calculate the velocity components using trigonometry

       Voy = vo without T

       Vox = Vo cost

       Voy = 7.3 sin 60

       Vox = 7.3 cos 60

       Voy = 6.32 m / s

       Vox = 3.65 m / s

Let's calculate the maximum height

         Y = 1.5 +6.32²/2 9.8

          Y = 3.54 m

This is the maximum height from the ground

b) They ask us for the position of this point horizontally, we can calculate it looking for the time it took for the skateboarder to reach the highest point

          Vfy = Voy - gt

          0 = Voy - gt

           t = Voy / g

           t = 6.32 / 9.8

           t = 0.645 s

Since there is no acceleration on the x-axis, we have a uniform movement, we can calculate the distance for this time

          X = Vox t

          X = 3.65 0.645

          X= 2.35 m