In an amusement park water slide, people slide down an essentially frictionless tube. The top of the slide is 3.1 m above the bottom where they exit the slide, moving horizontally, 1.4 m above a swimming pool.
What horizontal distance do they travel from the exit point before hitting the water?
Does the mass of the person make any difference?

Respuesta :

Answer:

  s = 4.167 m

Explanation:

given,

height of the slide = 3.1 m

horizontal movement above swimming pool = 1.4 m

using energy conservation

 [tex]m g h = \dfrac{1}{2} m v^2[/tex]

 [tex]v= \sqrt{2gh}}[/tex]

 [tex]v= \sqrt{2\times 9.8 \times 3.1}}[/tex]

 [tex]v= \sqrt{60.76}}[/tex]

      v = 7.79 m/s

using equation of motion

[tex]s = u t + \dfrac{1}{2}gt^2[/tex]

initial velocity is equal to zero

[tex]t = \dfrac{2s}{g}[/tex]

[tex]t = \sqrt{\dfrac{2\times 1.4}{9.8}}[/tex]

t = 0.535 s

Horizontal distance is equal to

 distance = speed x time

   s = 7.79 x 0.535

  s = 4.167 m

No, mass of the person will not make any difference.

The horizontal distance should be 4.167 m.

Also, it does not have any difference in the mass of the person.

Calculation of the horizontal distance:

here we used the conversation of energy should be used

[tex]mgh = \frac{1}{2} mv^2\\\\v = \sqrt{mgh}\\\\ = \sqrt{2\times 9.8\times 3.1}\\\\ = \sqrt{60.76}\\[/tex]

= 7.79 m/s

Since initial velocity should be equivalent to zero

So,

t = [tex]\sqrt{2\times 1.4 \div 9.8}[/tex]

= 0.535s

Now the distance is

= 0.779 (0.535)

= 4.167 m

Learn more about the mass here: https://brainly.com/question/25144785

Q&A Education