The drawing shows a skateboarder moving at v = 4.8 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is h = 0.48 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to which she rises above the end of the track.
To find the maximum height H to which the skateboarder rises above the end of the track, we can use the principles of projectile motion.
First, let's break down the initial velocity v into its horizontal and vertical components. The horizontal component (v_x) remains constant throughout the motion, while the vertical component (v_y) changes due to the gravitational acceleration.
v_x = v * cos(48°) (horizontal component)
v_y = v * sin(48°) (vertical component)
Given:
v = 4.8 m/s
angle of inclination (θ) = 48°
height above the ground (h) = 0.48 m
gravitational acceleration (g) = 9.8 m/s^2
Next, let's calculate the time it takes for the skateboarder to reach the maximum height. At this point, the vertical velocity will become zero.
Using the equation of motion:
v_y = v0_y + g * t
Since v_y = 0 at the maximum height, we can solve for t:
0 = v_y + g * t
t = -v_y / g
Now, let's substitute the values and solve for t:
t = -v * sin(48°) / g
Next, let's calculate the time it takes for the skateboarder to reach the end of the track. At this point, the height will be zero.
Using the equation of motion:
h = v0_y * t + (1/2) * g * t^2
Since h = 0 at the end of the track, we can solve for t:
0 = v * sin(48°) * t + (1/2) * g * t^2
Now, let's substitute the values and solve for t:
0 = v * sin(48°) * t + (1/2) * 9.8 * t^2
Simplifying the equation:
4.9 * t^2 = -v * sin(48°) * t
Next, let's solve for t:
t = 0 (initial time when at the end of the track)
t = -v * sin(48°) / (4.9)
Now that we have the time taken to reach the maximum height (t), let's calculate the maximum height (H).
Using the equation of motion:
H = v0_y * t + (1/2) * g * t^2
Substituting the values:
H = v * sin(48°) * t + (1/2) * 9.8 * t^2
Simplifying the equation:
H = v * sin(48°) * (-v * sin(48°) / (4.9)) + (1/2) * 9.8 * (-v * sin(48°) / (4.9))^2
Calculating the value of H will give us the maximum height to which the skateboarder rises above the end of the track.
To find the maximum height, we can use the concepts of projectile motion.
First, let's break down the skateboarder's motion into two components: horizontal (x-direction) and vertical (y-direction).
In the x-direction, the skateboarder's velocity remains constant at 4.8 m/s, as there is no horizontal force acting on her.
In the y-direction, the skateboarder's motion can be treated as a projectile. We know that the initial vertical velocity is 0 when she leaves the track, and the acceleration due to gravity is acting vertically downward.
Now, let's find the time it takes for the skateboarder to reach the maximum height. We can use the following kinematic equation:
v_f = v_i + at
Since the final vertical velocity at the maximum height is 0, the equation becomes:
0 = 0 + (-9.8 m/s^2) * t
Simplifying this equation, we find:
t = 0
This means that at the highest point, the skateboarder has spent 0 seconds in the air vertically.
Next, we can find the time it takes for the skateboarder to reach the end of the track horizontally. We can use the following kinematic equation:
d = v_i * t + 0.5 * a * t^2
where d is the horizontal distance traveled, v_i is the initial horizontal velocity (4.8 m/s), t is the time, and a is the horizontal acceleration (0 m/s^2).
Since there is no horizontal acceleration, the equation simplifies to:
d = v_i * t
Rearranging the equation, we get:
t = d / v_i
Now, we can find the horizontal distance d. The horizontal distance traveled can be calculated using trigonometric functions with the angle of 48° and the height h:
d = h / tan(48°)
Plugging in the given values:
d = 0.48 m / tan(48°)
Now, we can substitute this value for d in our equation for time:
t = (0.48 m / tan(48°)) / 4.8 m/s
Simplifying this equation, we find:
t = 0.107 s
Finally, to find the maximum height H, we can use the equation for vertical displacement:
d = v_i * t + 0.5 * a * t^2
Since the final vertical displacement at the maximum height is H, the equation becomes:
H = 0 + 0.5 * (-9.8 m/s^2) * (0.107 s)^2
Simplifying this equation, we find:
H = 0.058 m
Therefore, the maximum height to which the skateboarder rises above the end of the track is approximately 0.058 meters.