8.) A water slide whose top is 5.00 meters above the pool. The total length of t

Question

8.) A water slide whose top is 5.00 meters above the pool.   The total length of the slide is 15.0 meters because it ‘winds’ its way down to the water. Since water is running down the slide, there is very little friction (we will assume none) between the swimmers’ bathing suits and the slide surface.

a. Determine the final velocity of a person at the bottom of the slide before they enter the water.

Answer:______________ m/s

b. If the side were only 10.0 meters long (but at the same height) what would be the person’s final velocity? Note that the slide is steeper in this case.

Answer:_______________ m/s

c. If the person started at the top of the slide with an initial speed of 5.0 m/s, what would be their speed at the bottom of the slide for the 10-m long slide?

Answer:_____________ m/s

d. How long would it take the person to slide down for each of the 2 slides when they start from rest? For this problem, you may assume constant acceleration.

15.0 m slide: _________seconds

10.0 m slide: _________seconds

Explanation / Answer

as there is no friction, steeper surface would not have any role in finding final velocity (as energy is conserved)

though it may contribute in calculation of time.

a) mgh = 0.5mv2 => v = sqrt(2gh) = sqrt(2 x 9.8 x 5) = sqrt(98) = 9.8995 m/s

b)  mgh = 0.5mv2 => v = sqrt(2gh) = sqrt(2 x 9.8 x 5) = sqrt(98) = 9.8995 m/s

c) mgh + 0.5mvi2 = 0.5mv2 => v = sqrt(2gh + vi2) = sqrt(2 x 9.8 x 5 + 25) = sqrt(123) = 11.0905 m/s

d) in case of 15m, acceleration = g(5/15) = g/3 = 9.8/3 = 3.2666 m/s2

time = sqrt(2 x 15/3.26666) = 3.03 sec

incase of 10m, acceleration = g(5/10) = 4.9 m/s2

time = sqrt(2 x 10/4.9) = 2.0203 s

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