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Cliff divers at Acapulco jump into the sea from a cliff 37.1 m high. At the level of the sea, a rock sticks out a horizontal distance of 13.39 m. The acceleration of gravity is 9.8 m/s 2 . With what minimum horizontal velocity must the cliff divers leave the top of the cliff if they are to miss the rock? Answer in units of m/s.

Respuesta :

Answer:

[tex]v_x = 4.87 m/s[/tex]

Explanation:

Height of the cliff is given as

[tex]h = 37.1 m[/tex]

now the time taken by the diver to hit the surface is given as

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

[tex]37.1 = \frac{1}{2}(9.8)t^2[/tex]

[tex]t = 2.75 s[/tex]

Now in the same time it has to cover a distance of 13.39 m

so the speed in horizontal direction is given as

[tex]v_x = \frac{x}{t}[/tex]

[tex]v_x = \frac{13.39}{2.75}[/tex]

[tex]v_x = 4.87 m/s[/tex]

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