At an amusement park, a swimmer uses a water slide to enter the main pool. If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.28 m, what is her speed at the bottom of the slide?
Use the fact that the kinetic energy gained (1/2)MV^2, will equal the gravitational potential energy lost, MgH.
Therefore V^2 = 2 g H
H is the vertical height and g is the acceleration of gravity.
Solve for V. It will be in m/s if g is in m/s^2
To find the speed of the swimmer at the bottom of the slide, you can use the principle of conservation of mechanical energy. The total mechanical energy of the swimmer remains constant throughout the slide.
The total mechanical energy (E) is the sum of the kinetic energy (KE) and the potential energy (PE) at any given point.
E = KE + PE
The kinetic energy of an object is given by the equation:
KE = (1/2)mv^2
Where m is the mass of the object and v is its velocity.
At the top of the slide, the swimmer is at rest, so the initial kinetic energy is zero. The initial potential energy is given by:
PE_initial = mgh
Where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the vertical height of the slide (2.28 m).
Since there is no friction, you can assume that the only form of energy is the potential energy at the top of the slide, which is converted entirely into kinetic energy at the bottom of the slide.
Therefore, the total mechanical energy at the bottom of the slide is equal to the initial potential energy:
E = PE_final = mgh
Setting the initial potential energy equal to the final kinetic energy:
mgh = (1/2)mv^2
Simplify the equation by canceling out the mass:
gh = (1/2)v^2
Rearrange the equation to solve for v:
v^2 = 2gh
Take the square root of both sides to find v:
v = √(2gh)
Now you can calculate the speed of the swimmer at the bottom of the slide using the given values for g (9.8 m/s^2) and h (2.28 m).