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Cliff divers at Acapulco jump into the sea macias (fjm793) – Homework 3, 2d motion 19-20 – dowd – (WoffordWPHY11920 2) 4 from a cliff 36.2 m high. At the level of the sea, a rock sticks out a horizontal distance of 11.98 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:

v = 7.67 m/s

Explanation:

Given data:

horizontal distance 11.98 m

Acceleration due to gravity 9.8 m/s^2

Assuming initial velocity is zero

we know that

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

solving for t

we have

[tex]t = \sqrt{\frac{2h}{g}}[/tex]

substituing all value for time t

[tex]t = \sqrt{\frac{2\times 11.98}{9.8}}[/tex]

t = 1.56 s

we know that speed is given as

[tex]v = \frac{d}{t}[/tex]

[tex]v =\frac{11.98}{1.56}[/tex]

v = 7.67 m/s

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