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n amusement park water slide, people slide down an essentially frictionless tube. The top of the slide is 3.0m above the bottom where they exitthe slide, moving horizontally, 1.2mabove a swimming pool. What horizontal distance do they travel from the exit point before hitting the water

Respuesta :

Answer:

Horizontal distance will be 3.79 m

Explanation:

We have given h = 3 m

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

From conservation of energy we know that [tex]mgh=\frac{1}{2}mv^2[/tex]

[tex]v=\sqrt{2gh}[/tex]

[tex]v=\sqrt{2\times 9.8\times 3}=7.66m/sec[/tex]

Now initial vertical velocity = 0 m/sec

Initial horizontal velocity = 7.66 m/sec

Now height h = 1.2 m

From second equation of motion

[tex]h=ut+\frac{1}{2}gt^2[/tex]

[tex]1.2=0\times t+\frac{1}{2}\times 9.8\times t^2[/tex]

t = 0.494 sec

Now horizontal distance = Horizontal speed × time = 7.66×0.494 = 3.79 m

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