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
(a) How high above the ground is the highest point that the skateboarder reaches?
Well, if the skateboarder is shooting off a ramp, we can safely say they must be pretty high on life. But let's focus on the physics here. We can use some mathematical clownery to find the answer.
First, let's break down the initial velocity into its vertical and horizontal components. The vertical component can be found by multiplying the initial velocity (6.5 m/s) by the sine of the angle (62°):
Vertical component = 6.5 m/s * sin(62°)
Now, we know that at the highest point, the vertical velocity will be zero. So, using the vertical component of the initial velocity, we can find how long it takes to reach this point using the formula:
Time = (Vertical component) / (acceleration due to gravity)
Given that the acceleration due to gravity is approximately 9.8 m/s^2 (and gravity is always trying to bring us down, just like a bad pun), we can calculate the time it takes to reach the highest point.
Next, we can find the vertical distance traveled by the skateboarder using the formula:
Vertical distance = (Vertical component)^2 / (2 * acceleration due to gravity)
Don't worry, we're almost there!
Finally, to find the height above the ground at the highest point, we can subtract the initial height of the ramp (1.5 m) from the vertical distance traveled by the skateboarder.
(b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
To answer this question, we just need to find the horizontal distance traveled by the skateboarder using the horizontal component of the initial velocity. This can be calculated by multiplying the initial velocity (6.5 m/s) by the cosine of the angle (62°).
Horizontal distance = 6.5 m/s * cos(62°)
And there you have it! With these calculations, you should be able to find the answers you're looking for. Just remember, even if the skateboarder doesn't go higher than your expectations, they're still undoubtedly having a wheely good time!
To solve this problem, we can break it down into two components: horizontal and vertical.
(a) To find the highest point reached by the skateboarder, we need to determine the vertical component of the velocity at that point. We can use the following equation:
Vy = V * sin(θ)
where Vy is the vertical component of velocity, V is the initial velocity (6.5 m/s), and θ is the angle of projection (62°).
Vy = 6.5 * sin(62°)
Vy ≈ 5.69 m/s
Next, we need to find the time it takes for the skateboarder to reach the highest point. We can use the equation:
Vy = g * t
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time.
5.69 = 9.8 * t
t ≈ 0.58 s
Now, we can find the height above the ground at the highest point using the equation:
h = h0 + Vy * t - 0.5 * g * t^2
where h0 is the initial height above the ground (1.5 m), Vy is the vertical component of velocity (5.69 m/s), t is the time (0.58 s), and g is the acceleration due to gravity (9.8 m/s^2).
h = 1.5 + 5.69 * 0.58 - 0.5 * 9.8 * (0.58)^2
h ≈ 3.16 m
Therefore, the highest point reached by the skateboarder is approximately 3.16 m above the ground.
(b) To find the horizontal distance from the end of the ramp to the highest point, we can use the equation:
Sx = V * cos(θ) * t
where Sx is the horizontal distance, V is the initial velocity (6.5 m/s), θ is the angle of projection (62°), and t is the time (0.58 s).
Sx = 6.5 * cos(62°) * 0.58
Sx ≈ 1.68 m
Therefore, when the skateboarder reaches the highest point, the horizontal distance from the end of the ramp is approximately 1.68 m.
To solve this problem, we can break it down into two parts: finding the maximum height reached (part a) and finding the horizontal distance from the end of the ramp to the highest point (part b).
(a) To find the highest point reached by the skateboarder, we can use the projectile motion equations. The initial vertical velocity (Vy) can be found using the given velocity and the angle of projection.
Vy = Velocity * sin(Angle)
Vy = 6.5 m/s * sin(62°)
Next, we can use this vertical velocity to find the time it takes for the skateboarder to reach the highest point, assuming no air resistance. We can use the formula for time of flight (T) for a projectile motion:
T = (2 * Vy) / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the value of Vy we determined earlier:
T = (2 * (6.5 m/s * sin(62°))) / 9.8 m/s^2
Once we have the time of flight, we can find the maximum height (H) reached using the formula:
H = Vy*T - (1/2) * g * T^2
Substituting the values of Vy and T:
H = (6.5 m/s * sin(62°)) * ((2 * (6.5 m/s * sin(62°))) / 9.8 m/s^2) - (1/2) * 9.8 m/s^2 * ((2 * (6.5 m/s * sin(62°))) / 9.8 m/s^2)^2
Simplifying this equation will give us the answer for part (a).
(b) To find the horizontal distance from the end of the ramp to the highest point, we need to find the horizontal velocity (Vx). The horizontal velocity remains constant throughout the motion and can be found using the equation:
Vx = Velocity * cos(Angle)
Vx = 6.5 m/s * cos(62°)
Now, using the time of flight (T) calculated earlier, we can find the horizontal distance (D) traveled using the equation:
D = Vx * T
Substituting the values of Vx and T will give us the answer for part (b).
Note: Make sure to convert the angles to radians if your calculator uses radians instead of degrees.