The drawing shows a skateboarder moving at v = 5 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is h = 0.52 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.
Isn't this best answered by energy relationships?
Launch velocity upward= 5sinTheta
KE of the horizontal motion is constant, so the change in PE is due to the vertical KE.
mgh=1/2 m 25 sin^2Theta.
This is the h above the track the skateboarder rises .
7.45
Yes, you're right! This problem can be solved using energy relationships. Let's go step by step to find the maximum height H to which the skateboarder rises above the end of the track.
1. Firstly, let's find the launch velocity upward. The skateboarder is moving horizontally at a velocity v = 5 m/s. The upward velocity can be found using the angle of the track slant: v_upward = v * sin(θ), where θ = 48°.
2. The kinetic energy (KE) of the horizontal motion is constant, so the change in potential energy (PE) is due to the vertical KE. We can equate the initial gravitational potential energy (mgh) at the end of the track to the final vertical kinetic energy (1/2 m v_upward^2) at the maximum height H.
mgh = 1/2 m v_upward^2
3. Now, substitute the value of v_upward (v * sin(θ)) and solve the equation for H.
mgh = 1/2 m (v * sin(θ))^2
gh = 1/2 (v^2) (sin(θ))^2
H = (v^2) (sin(θ))^2 / (2g)
Here, g is the acceleration due to gravity (approximately 9.8 m/s^2).
4. Plug in the given values of v = 5 m/s, θ = 48°, and g = 9.8 m/s^2, into the equation to calculate the maximum height H.
H = (5^2) (sin(48°))^2 / (2 * 9.8)
5. Solve the equation to find the maximum height H.
H ≈ 1.471 m
Therefore, the maximum height to which the skateboarder rises above the end of the track is approximately 1.471 meters.