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.

Start at the bottom of the track. Her verical component of velocity is 5sinTheta. So the KE which drives the vertical rise is 1/2 m 25sin^2 theta.
Set that equal to the change in potential energy, mgh. Solve for h , the height above GROUND. To get the heighht above the end of track,subtract 0.52m.

I'm sorry I'm not very good at this but can you please clarify:

1. 5sinTheta - are you saying the sign of 5 degrees and what do I do with it?

2. 1/2m25sin^2theta - are you saying 1/2mv(squared) then find the sine of v (squared)?

Thanks!

"5 sin theta" means 5 x sin 48 = 3.716
"theta" is the slope angle.

"1/2m25sin^2theta " means
(1/2)x m x 25 x (sin 38)^2
(The 5-squared became the 25)
You don't need to know tha mass m to solve the problem, because it will cancel out when potential energy m g h is set equal to (1/2)x m x 25 x (sin 38)^2.

Well, it seems like you've got the right idea! Let me help you clarify a little further:

1. The value "5" in "5sinTheta" represents the initial velocity of the skateboarder, which is given as 5 m/s. The "sinTheta" term is the sine of the angle theta, which is 48 degrees in this case. So, you'll calculate 5 x sin(48) to find the vertical component of the velocity.

2. The expression "1/2m25sin^2theta" is trying to represent the kinetic energy (KE) of the skateboarder in terms of the given values. Let's break it down:
- "1/2mv^2" refers to the standard formula for kinetic energy, where "m" represents the mass and "v" is the velocity. In this case, you don't need to know the mass of the skateboarder, as it will cancel out later in the equation.
- The value "25" comes from squaring the initial velocity (5 squared).
- "sin^2theta" refers to the square of the sine of the angle theta.

To sum it up, you need to calculate 1/2 x m x (5^2) x (sin^2(48)) for the kinetic energy, and set it equal to mgh for the potential energy, where "h" represents the height above the ground. Don't forget to subtract 0.52m to find the height above the end of the track.

I apologize for the confusion in my previous response. There seems to be a typographical error in the calculations. Let's go through the correct steps to find the maximum height H.

1. First, let's find the initial vertical component of velocity, which is the product of the velocity (v = 5 m/s) and the sine of the angle (θ = 48°). The correct calculation is:

Vertical component of velocity = v × sin(θ) = 5 m/s × sin(48°) ≈ 3.799 m/s

2. The kinetic energy (KE) that drives the vertical rise is equal to 1/2 × m × (vertical component of velocity)^2. However, mass (m) is not necessary to find the maximum height and will cancel out. The correct calculation is:

KE = 1/2 × m × (vertical component of velocity)^2 = 1/2 × 3.799 m/s × 3.799 m/s ≈ 7.225 J

3. The change in potential energy (ΔPE) is equal to the mass (m) × gravitational acceleration (g) × height (h). Ignore the mass, as it cancels out. The correct calculation is:

ΔPE = m × g × h = 9.8 m/s² × h

4. Setting the kinetic energy equal to the change in potential energy, we have:

KE = ΔPE
7.225 J = 9.8 m/s² × h
h ≈ 0.738 m

5. Finally, to find the maximum height above the end of the track, subtract the initial height of the track above the ground. The correct calculation is:

H = h - 0.52 m ≈ 0.738 m - 0.52 m ≈ 0.218 m

Therefore, the skateboarder rises to a maximum height of approximately 0.218 meters 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 follow these steps:

1. Calculate the vertical component of velocity: Given that the skateboarder is moving at a velocity v = 5 m/s, and the track is slanted upward by 48°, we can calculate the vertical component of velocity using the formula v_vertical = v × sin(theta). Here, theta represents the slope angle, which is 48°. So, v_vertical = 5 × sin(48°) = 3.716 m/s.

2. Calculate the kinetic energy driving the vertical rise: The kinetic energy (KE) driving the vertical rise is given by the formula KE = (1/2) × m × v_vertical^2, where m is the mass of the skateboarder (which we can ignore since it cancels out later). Substituting in the values, KE = (1/2) × 25 × (3.716)^2 = 86.11 J.

3. Set the kinetic energy equal to the change in potential energy: At the maximum height H, the entire kinetic energy is converted to potential energy. So, we can equate KE to the change in potential energy, which is given by the formula m × g × h, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height above the ground. Thus, we have 86.11 J = 25 × 9.8 × h.

4. Solve for h, the height above the ground: Rearranging the equation, we find h = 86.11 J / (25 × 9.8) = 0.352 m.

5. Find the height above the end of the track: Since the height given in the problem is h = 0.52 m above the ground, we subtract this height from the height above the ground we just found: H = 0.352 m - 0.52 m = -0.168 m.

The maximum height above the end of the track to which the skateboarder rises is -0.168 m. Note that the negative sign indicates that the skateboarder falls below the end of the track.