Question

In an amusement park water slide, people slide down an essentially frictionless tube. The top of the slide is 3.2 m above the bottom where they exit the slide, moving horizontally, 1.5 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?

Answer 1

Answer:

Horizontal distance moved by the person is 3.2 m

And it is independent of the mas of the person

Explanation:

As we know that the speed of the people when it is sliding downwards is given as

$\frac{1}{2}mv^2 = mg(h_2 - h_1)$

$v = \sqrt{2g(h_2 - h_1)}$

$v = \sqrt{2(9.81)(3.2 - 1.5)}$

$v = 5.77 m/s$

Now the time taken by the person to hit the water surface is given as

$\frac{1}{2}gt^2 = h$

$\frac{1}{2}(9.81)t^2 = 1.5$

$t = 0.55 s$

horizontal distance moved by the person is given as

$d = vt$

$d = 5.77 (0.55)$

$d = 3.2 m$