Starting from rest at the top, a child slides down the water slide at a swimming pool and enters the water at a final speed of 5.59 m/s. At what final speed would the child enter the water if the water slide were twice as high? Ignore friction and resistance from the air and the water lubricating the slide.
(1/2) m v^2 = m g h
(1/2) m u^2 = 2 m g h
u/v = sqrt 2
so
5.59 * 1.414
the kids swimming in the pool gravity or friction.
To solve this problem, we can use the principle of conservation of mechanical energy. According to this principle, the initial mechanical energy of the child at the top of the slide is equal to the final mechanical energy of the child in the water.
The mechanical energy of an object can be expressed as the sum of its potential energy (PE) and kinetic energy (KE). At the top of the slide, the child only has potential energy, and in the water, the child only has kinetic energy.
The potential energy (PE) of the child can be calculated using the equation:
PE = m * g * h
Where:
- m is the mass of the child
- g is the acceleration due to gravity (9.8 m/s^2)
- h is the height of the slide
Since the child starts from rest, the initial kinetic energy (KE) is zero. The final kinetic energy can be calculated using the equation:
KE = 0.5 * m * v^2
Where:
- m is the mass of the child
- v is the final velocity of the child in the water
Since the problem states that the child enters the water at a final speed of 5.59 m/s for the original slide height, we can set up the equation:
PE = KE
m * g * h = 0.5 * m * v^2
The mass of the child cancels out, leaving:
g * h = 0.5 * v^2
Now, let's solve for the final speed of the child (v) for the double slide height.
Since the height of the slide is doubled, the new height will be 2h. So, the equation becomes:
g * (2h) = 0.5 * v_new^2
Substituting the known values:
9.8 * (2h) = 0.5 * v_new^2
Simplifying:
19.6h = 0.25v_new^2
Dividing both sides by 0.25:
78.4h = v_new^2
Taking the square root of both sides:
v_new = sqrt(78.4h)
Therefore, if the height of the water slide is doubled, the child would enter the water at a final speed of sqrt(78.4h).
To solve this problem, we can use the principles of conservation of energy, specifically the conservation of mechanical energy. The initial potential energy at the top of the slide is converted into kinetic energy at the bottom, assuming no energy losses due to friction or air resistance.
Let's assume the initial height of the water slide is H, and the final speed is V. We are given that the final speed V is 5.59 m/s for the original height.
The kinetic energy of the child at the bottom of the slide is given by:
K.E = 1/2 * m * V^2
The potential energy at the top of the slide is given by:
P.E = m * g * H
where m is the mass of the child, g is the acceleration due to gravity (approximately 9.8 m/s^2).
We can set the initial potential energy equal to the final kinetic energy:
m * g * H = 1/2 * m * V^2
Now, let's consider the case when the water slide is twice as high. Let's call the new height 2H. Since no energy is lost due to friction or air resistance, the child will have the same kinetic energy at the bottom. Therefore, we can write:
m * g * (2H) = 1/2 * m * V^2
Simplifying the equation, we have:
2 * (m * g * H) = 1/2 * m * V^2
Since m and g are constant, we can cancel them out:
2 * H = 1/2 * V^2
Solving for V, we get:
V^2 = 4 * H
Taking the square root of both sides, we find:
V = 2 * sqrt(H)
Therefore, if the water slide were twice as high, the child would enter the water with a final speed of 2 * sqrt(H).
Note: It's important to remember that this solution assumes no energy losses due to friction or air resistance. In real-world situations, these factors may affect the final speed of the child.