A skateboarder shoots off a ramp with a velocity of 6.7 m/s, directed at an angle of 59° above the horizontal. The end of the ramp is 1.1 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?
I have looked at several examples and even answers for this question but I cannot seem to understand what I need to do. Can someone please explain this question to me? Thanks!
check the related questions. The one with the 6.8m/s
Certainly! Let's break down the problem step by step.
(a) To find the height above the ground that the skateboarder reaches, we need to analyze the vertical component of their motion.
First, let's determine the initial vertical velocity (vy) of the skateboarder. We know that the skateboarder shoots off the ramp with a velocity of 6.7 m/s, directed at an angle of 59° above the horizontal. To find the vertical component, we use the formula vy = v * sin(θ), where v is the initial velocity (6.7 m/s) and θ is the angle (59°).
vy = 6.7 m/s * sin(59°)
vy ≈ 6.7 m/s * 0.8575
vy ≈ 5.74 m/s
Next, we can use the kinematic equation for vertical motion to find the maximum height (h) reached by the skateboarder. The equation is: vy^2 = v0y^2 + 2 * a * h, where v0y is the initial vertical velocity, a is the acceleration (due to gravity), and h is the maximum height.
Since the skateboarder is at the highest point, the final vertical velocity (vf) is 0 m/s. So, we can rearrange the equation as follows:
0 = (5.74 m/s)^2 - 2 * 9.8 m/s^2 * h
Solving for h:
2 * 9.8 m/s^2 * h = (5.74 m/s)^2
h = ((5.74 m/s)^2) / (2 * 9.8 m/s^2)
h ≈ 1.765 m
Therefore, the highest point reached by the skateboarder above the ground is approximately 1.765 meters.
(b) To find the horizontal distance (x) from the end of the ramp to the highest point, we can analyze the horizontal component of the skateboarder's motion.
The horizontal velocity (vx) remains constant throughout the motion. We can find it using the formula vx = v * cos(θ), where v is the initial velocity (6.7 m/s) and θ is the angle (59°).
vx = 6.7 m/s * cos(59°)
vx ≈ 6.7 m/s * 0.5141
vx ≈ 3.44 m/s
Now, we can calculate the time it takes for the skateboarder to reach the highest point by using the equation vy = v0y + a * t, where vy is the vertical velocity, v0y is the initial vertical velocity, a is the acceleration, and t is the time.
Since the skateboarder reaches the highest point when vy is 0 m/s, we have:
0 = 5.74 m/s - 9.8 m/s^2 * t
Solving for t:
9.8 m/s^2 * t = 5.74 m/s
t ≈ 0.585 s
Finally, we can calculate the horizontal distance by multiplying the horizontal velocity by the time:
x = vx * t
x ≈ 3.44 m/s * 0.585 s
x ≈ 2.01 m
Therefore, when the skateboarder reaches the highest point, they are approximately 2.01 meters horizontally away from the end of the ramp.
I hope this explanation helps! Let me know if you have any further questions.