A skateboarder shoots off a ramp with a velocity of 6.5 m/s, directed at an angle of 63° 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?

(a) The vertical rise H can be determined by requiring that the vertical portion of the kinetic energy, (1/2)M*(V sin 63)^2, be converted to potential energy M g H.

H = (1/2)(V sin63)^2/g
Add H to the initial platform height, 1.5 m, for the highest point elevation above ground.
(b) Multiply the time to reach maximum height (zero vetical velocity), (Vsin63)/g, by the horizontal velocity component, V cos63.

X = (V^2/g)sin63 cos63

To solve both parts of the problem, we can break down the skateboarder's motion into horizontal and vertical components. Let's start with part (a).

(a) To determine how high above the ground the skateboarder reaches, we need to find the maximum height of the projectile's trajectory.

The initial velocity of the skateboarder is given as 6.5 m/s at an angle of 63° above the horizontal. We can decompose this velocity into its horizontal and vertical components.

The horizontal component (Vx) can be calculated using the equation Vx = V * cos(θ), where V represents the initial velocity and θ represents the angle above the horizontal. Let's plug in the values and calculate it.

Vx = 6.5 m/s * cos(63°) = 6.5 m/s * 0.447 = 2.91 m/s (rounded to two decimal places)

Next, we need to find the vertical component (Vy) using the equation Vy = V * sin(θ).

Vy = 6.5 m/s * sin(63°) = 6.5 m/s * 0.894 = 5.84 m/s (rounded to two decimal places)

At the highest point, the vertical velocity component (Vy) will be zero, as the skateboarder momentarily stops moving upward before descending.

Using the equation Vy^2 = V0y^2 + 2 * a * Δy, where Vy represents the final vertical velocity, V0y represents the initial vertical velocity, a represents the acceleration due to gravity (-9.8 m/s^2), and Δy represents the change in height.

Since Vy is zero, we can rewrite the equation as 0 = (5.84 m/s)^2 + 2 * (-9.8 m/s^2) * Δy.

Now we can solve for Δy, the change in height.

(5.84 m/s)^2 = 2 * (-9.8 m/s^2) * Δy

34.06 m^2/s^2 = -19.6 m/s^2 * Δy

Δy = 34.06 m^2/s^2 / (-19.6 m/s^2) = -1.74 m

Since we are looking for the height above the ground, Δy is negative because it is below the ramp.

Therefore, the highest point above the ground that the skateboarder reaches is 1.5 m - 1.74 m = -0.24 m (rounded to two decimal places).

(b) To determine the horizontal distance from the end of the ramp to the highest point, we can use the equation Δx = Vx * t, where Δx represents the horizontal displacement and t represents the time it takes for the skateboarder to reach the highest point.

To find t, we can use the equation Vy = V0y + a * t, where Vy represents the final vertical velocity, V0y represents the initial vertical velocity, a represents the acceleration due to gravity (-9.8 m/s^2), and t represents the time.

Since Vy is zero at the highest point, we can rewrite the equation as 0 = 5.84 m/s + (-9.8 m/s^2) * t.

Solving for t:

5.84 m/s = (-9.8 m/s^2) * t

t = 5.84 m/s / (-9.8 m/s^2) = -0.595 s (rounded to three decimal places)

Since we are only concerned with the magnitude of time, t is positive. Therefore, t = 0.595 s.

Now we can calculate Δx using the horizontal component Vx and time t.

Δx = 2.91 m/s * 0.595 s = 1.73 m (rounded to two decimal places)

Therefore, when the skateboarder reaches the highest point, the horizontal distance from the end of the ramp is approximately 1.73 meters.