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21 April, 08:26

Cliff divers at Acapulco jump into the sea

from a cliff 31.5 m high. At the level of the

sea, a rock sticks out a horizontal distance of

12.49 m.

The acceleration of gravity is 9.8 m/s.

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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Answers (1)
  1. 21 April, 10:07
    0
    V=4.927 m/s

    Explanation:

    H=Vi*t+0.5*g*t² ... 1

    initial velocity is Vi=0 m/s

    H=31.5

    putting value of H in equation 1

    31.5=0+0.5*9.81*t²

    t=2.535 s

    S=V*t ... 2

    Horizontal distance S=12.49 m

    12.49=V * (2.535)

    V=4.927 m/s
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