A skateboarder shoots off a ramp with a velocity of 6.5 m/s, directed at an angle of 62° 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?
m

(b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
m

To solve this problem, we can break down the initial velocity of the skateboarder into horizontal and vertical components.

Let's start by finding the initial vertical velocity (Viy). We can use the equation:
Viy = V * sin(θ)

where V is the initial velocity (6.5 m/s) and θ is the angle of the ramp (62°).

Viy = 6.5 m/s * sin(62°)
Viy = 6.5 m/s * 0.88
Viy ≈ 5.72 m/s

Now, let's find the time it takes for the skateboarder to reach the highest point of his trajectory. We can use the equation for vertical displacement:
Δy = Viy * t - 0.5 * g * t^2

where Δy is the vertical displacement, Viy is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (9.8 m/s^2).

At the highest point, the vertical displacement is equal to the ramp height (1.5 m). Therefore, we can rewrite the equation as:

1.5 m = 5.72 m/s * t - 0.5 * 9.8 m/s^2 * t^2

Rearranging this equation gives us a quadratic equation in terms of t:

0.5 * 9.8 m/s^2 * t^2 - 5.72 m/s * t + 1.5 m = 0

Solving this quadratic equation we find two values for t. We select the positive value because the negative value represents a time before the skateboarder left the ramp.

The highest point is reached when t = 0.67 s (rounded to two decimal places).

Now, let's find the vertical displacement at this highest point. We can use the same equation mentioned earlier:

Δy = Viy * t - 0.5 * g * t^2

Δy = 5.72 m/s * 0.67 s - 0.5 * 9.8 m/s^2 * (0.67 s)^2
Δy ≈ 1.67 m

Therefore, the highest point that the skateboarder reaches above the ground is approximately 1.67 meters.

Moving on to part (b), we need to find the horizontal distance (Δx) from the end of the ramp to the highest point. We can use the equation for horizontal displacement:

Δx = V * cos(θ) * t

where V is the initial velocity (6.5 m/s), θ is the angle of the ramp (62°), and t is the time it takes for the skateboarder to reach the highest point (0.67 s).

Δx = 6.5 m/s * cos(62°) * 0.67 s
Δx ≈ 3.39 m

Therefore, the highest point horizontally from the end of the ramp is approximately 3.39 meters.

To summarize:
(a) The highest point above the ground that the skateboarder reaches is approximately 1.67 meters.
(b) The highest point is approximately 3.39 meters horizontally from the end of the ramp.