In an amusement park, there is a water slide that starts h meters above its base where people slide out 1.2 m above the surface of a pool. You may assume there is essentially no friction between a person and the slide.
a) What should be the maximum (safe) height h of the slide if the length of the pool is 3.0 m?
b) Does the mass of the person matter?
c) What should be the maximum (safe) height h if the slide is 0.6 m above the pool?
a person should land in the pool (as opposed to passing over it)
... their velocity is such that it allows the fall from 1.2 m before they clear the 3.0 m length
find the time to free-fall 1.2 m
... divide the time into 3.0 m to find the max safe velocity
... calculate the max safe slide height ... h = v^2 / (2 g)
mass is not used in any of the calculations
use the same technique for the 0.6 m drop
when dividing the time by 3.0, wouldnt the answer be in s/m, which is not the correct velocity term. And when dividing 3.0 by the time, it comes up as 6 m/s which seems a little too high. Am I wrong here?
I got 0.49 seconds for the time.
he said to divide the time into 3.0m
That is, divide the 3.0m by time, giving m/s
To answer these questions, we need to consider the principles of potential energy and kinetic energy.
a) First, let's determine the maximum safe height of the slide when the pool length is 3.0 m. The key concept here is conservation of energy. When the person reaches the bottom of the slide, their potential energy is converted into kinetic energy, which is then further converted to potential energy as they reach the surface of the pool.
The potential energy of an object at a height h is given by the equation: Ep = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
The kinetic energy of an object moving at a velocity v is given by the equation: Ek = (1/2)mv^2, where m is the mass of the object and v is the velocity.
Since we are assuming there is essentially no friction, the final velocity of the person sliding into the pool will be zero (since all the kinetic energy will be converted to potential energy again). Therefore, we can equate the potential energy at the top of the slide to the potential energy at the point where they are 1.2 m above the surface of the pool:
mgh = mgh' + (1/2)mv^2
where h' is the height above the surface of the pool (1.2 m in this case) and v is the final velocity (which is zero).
h = h' + (1/2)v^2/g
Since v^2 and g remain constant, we can rearrange the equation to find:
h = h' + (1/2)v^2/g
Substituting the values of h' = 1.2 m and g = 9.8 m/s^2, we get:
h = 1.2 + 0 = 1.2 m
Therefore, the maximum safe height of the slide should be 1.2 m when the pool length is 3.0 m.
b) The mass of the person does not matter in determining the safe height of the slide. This is because the mass cancels out when equating the potential energy at the top of the slide to the potential energy at the point where they are 1.2 m above the surface of the pool. Therefore, the height of the slide remains the same regardless of the person's mass.
c) To calculate the maximum safe height of the slide when it is 0.6 m above the pool, we can use the same equation as in part (a):
h = h' + (1/2)v^2/g
Now, substituting the values h' = 0.6 m and g = 9.8 m/s^2, we get:
h = 0.6 + 0 = 0.6 m
Therefore, the maximum safe height of the slide should be 0.6 m when it is 0.6 m above the pool.