Question #95677

Given: The battleship and enemy ships A
and B lie along a straight line. Neglect air
friction.
A battleship simultaneously fires two shells
at these two enemy ships.from a cliff 33.8 m high. At the level of the
sea, a rock sticks out a horizontal distance of
8.29 m. The acceleration of gravity is 9.8 m/s2. 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.

Expert's answer

The time of flight:


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

The minimum horizontal velocity:


v=dt=dg2hv=\frac{d}{t}=d\sqrt{\frac{g}{2h}}

v=8.299.82(33.8)=3.16msv=8.29\sqrt{\frac{9.8}{2(33.8)}}=3.16\frac{m}{s}


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