A skateboarder shoots off a ramp with a velocity of 6.8 m/s, directed at an angle of 54° above the horizontal. The end of the ramp is 1.3 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?
vertical problem first:
Vi = 6.8 sin 54
v = Vi - g t
at top 0 = Vi - g t
so t at top = Vi/g
y at top = 1.3 + Vi t - 4.9 t^2
horizontal problem
u = 6.8 cos 54
x = u t
To determine the answers, we can break down the given information and solve step by step using the principles of projectile motion.
(a) The highest point that the skateboarder reaches is when the vertical component of the velocity becomes zero. This occurs when the skateboarder reaches the peak of their trajectory.
To find this height, we need to use the following formula:
Vf² = Vi² + 2ad
Where:
Vf = Final velocity in the y direction
Vi = Initial velocity in the y direction
a = Acceleration in the y direction (gravity)
d = Vertical displacement
In this case, it's given that Vi = 6.8 m/s and a = -9.8 m/s² (negative because it's directed downward). We are looking for d, the vertical displacement.
As the skateboarder reaches the highest point, Vf = 0, and the equation becomes:
0 = 6.8² - 2(-9.8)d
Solving for d, we have:
39.04 = 19.6d
d = 39.04 / 19.6
d ≈ 2 meters
Therefore, the highest point that the skateboarder reaches above the ground is approximately 2 meters.
(b) To find the horizontal distance from the end of the ramp to the highest point, we can use the horizontal component of the velocity.
The horizontal component of the velocity does not change throughout the motion. Let's call this Vx.
To find Vx, we can use the formula:
Vx = Vi * cosθ
Where:
Vi = Initial velocity
θ = Angle of projection
In this case, Vi = 6.8 m/s and θ = 54°.
Vx = 6.8 * cos(54°)
Vx ≈ 6.8 * 0.5878
Vx ≈ 3.992 meters/second
Since the horizontal distance traveled can be determined using the time, we need to find the time it takes for the skateboarder to reach the highest point.
The formula to find the time of flight is:
T = (2 * Vi * sinθ) / a
Where:
T = Time of flight
Vi = Initial velocity
θ = Angle of projection
a = Acceleration in the y direction (gravity)
In this case,
T = (2 * 6.8 * sin(54°)) / 9.8
T ≈ 1.367 seconds
Now we can calculate the horizontal distance (x) using the formula:
x = Vx * T
x = 3.992 * 1.367
x ≈ 5.465 meters
Therefore, when the skateboarder reaches the highest point, the horizontal distance from the end of the ramp is approximately 5.465 meters.