Physics: 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.

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Cliff divers at acapulco jump into the sea macias

MrKrinkle77

08.01.2022

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

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

solving for t

we have

$t = \sqrt{\frac{2h}{g}}$

substituing all value for time t

$t = \sqrt{\frac{2\times 11.98}{9.8}}$

t = 1.56 s

we know that speed is given as

$v = \frac{d}{t}$

$v =\frac{11.98}{1.56}$

v = 7.67 m/s

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