Velocity & Horizontal Distance of Person on Water Slide

[chunky]

a person slides down a water slide that is 61 m long and has a slope angle of 24 degrees and the end is a ramp with a height of 3.66 m with an angle of 30 degrees to the horizontal determine the magnitude and direction of the velocity of the person when they just fly away from the ramp and the horizontal distance that they fly.

View attachment 2500

 

[Ackbach]

There are a number of equations that I like to say "have physics in them", as opposed to being merely descriptive. Your equations with physics, in a typical order of presentation, are the following:

1. Newton's Second Law
2. Conservation of Energy
3. Work-Energy Theorem
4. Conservation of Linear Momentum
5. Conservation of Angular Momentum

Which of these approaches do you think would be appropriate for this problem?

 

[chunky]

I believe the most appropriate approach would be the conservation of energy ie,

T1+V1+U1-2=T2+V2

Then just find all of the relevant pieces of information from the question and subbing them into the equation and solving for the magnitude and direction of the velocity and well as the max flying distance

 

[Ackbach]

I would agree that CoE would be the right approach for the first part of this problem (finding the velocity). Once you do that, however, how would you continue?

 

[CellCoree]

I would first like to clarify that the velocity of the person on the water slide can be calculated using the principles of Newton's laws of motion and the laws of conservation of energy.

To determine the magnitude and direction of the velocity of the person when they fly away from the ramp, we can use the equation for conservation of energy, which states that the initial potential energy (mgh) is equal to the final kinetic energy (1/2mv^2) of the person. In this case, the initial potential energy is equal to the final kinetic energy when the person just flies away from the ramp.

Using this equation, we can calculate the velocity (v) of the person as follows:

mgh = 1/2mv^2

Where m is the mass of the person, g is the acceleration due to gravity (9.8 m/s^2), h is the height of the ramp (3.66 m), and the angle of the ramp (30 degrees) is the angle between the ramp and the horizontal.

Solving for v, we get:

v = √(2gh)

Substituting the values, we get:

v = √(2*9.8*3.66) = 8.26 m/s

Therefore, the magnitude of the velocity of the person when they fly away from the ramp is 8.26 m/s.

To determine the direction of the velocity, we can use the concept of vector addition. The person's velocity is a combination of two components: the horizontal component and the vertical component. The horizontal component is equal to the initial velocity of the person on the water slide, which is 0 m/s. The vertical component is equal to the velocity calculated above, which is 8.26 m/s.

Using the Pythagorean theorem, we can calculate the magnitude of the velocity vector:

|v| = √(0^2 + 8.26^2) = 8.26 m/s

To determine the direction, we can use the inverse tangent function:

θ = tan^-1(8.26/0) = 90 degrees

Therefore, the direction of the velocity of the person when they fly away from the ramp is 90 degrees, which is straight up.

To calculate the horizontal distance that the person flies, we can use the equation for horizontal distance, which is:

d = v*cosθ*t

Where d is the horizontal