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 = 0.54 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.
While rising 0.54m at the end of the track, her velocity decreases to V2 that is given by
V1^2 - V2^2 = 2gh = 10.58 m^2/s^2
where V1 = 4.8 m/s
That tells you that
V2 = sqrt(23.04-10.58) = 3.53 m/s
The vertical velocity component leaving the ramp is
V2y = V2*sin48 = 2.62 m/s
How high she rises depends upon the value of V2y.
2gH = (V2y)^2
Solve for H.
drwls response is the correct way to do this problem.
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 problem into two parts: the horizontal motion and the vertical motion.
Horizontal Motion:
The skateboarder is moving horizontally at a constant speed of v = 4.8 m/s. This means there is no acceleration in the horizontal direction. Therefore, the horizontal displacement is given by:
Δx = v * t
where Δx represents the horizontal distance traveled and t represents time.
Vertical Motion:
Since the skateboarder follows the characteristic path of projectile motion, we can consider only the vertical motion. The acceleration due to gravity acts vertically downward. We can use the following equation to calculate the maximum height H:
H = h + Δy
where h is the height of the end of the track above the ground and Δy is the vertical displacement of the skateboarder.
To find Δy, we need to determine the time it takes for the skateboarder to leave the track and reach the maximum height. This can be done using the equation of motion:
Δy = v0 * t - (1/2) * g * t^2
where v0 represents the initial vertical velocity, t represents time, and g is the acceleration due to gravity.
In this case, the skateboarder leaves the track with an initial vertical velocity of zero (since she follows the characteristic path of projectile motion). Therefore, the equation simplifies to:
Δy = - (1/2) * g * t^2
We can solve for t by equating the horizontal displacement Δx to the horizontal component of velocity (v) multiplied by the time (t):
Δx = v * t
Substituting the given values:
v = 4.8 m/s
Δx = 0.54 m
t = Δx / v
Now we can substitute the value of t into the equation for Δy and solve for Δy:
Δy = - (1/2) * g * (Δx / v)^2
Next, we can substitute the value of Δy into the equation for H and solve for H:
H = h + Δy
By plugging in the given values:
h = 0.54 m
Δy = - (1/2) * g * (Δx / v)^2
H = h + Δy
We can substitute the known values to calculate H.