Question

Divers in Acapulco dive from a cliff that is 65 m high. If the rocks below the cliff extend outward for 25 m, how fast must the diver be going when he jumps if he just barely clears the rocks ?

Answer 1

Given,

The height of the cliff, h=65 m

The extension length of the cliff, i.e., the range of the jump, R=25 m

From the equation of motion, the height of the jump is given by,

`h=(1)/(2)gt^2`

Where g is the acceleration due to gravity and t is the time of flight of the diver.

On substituting the known values,

`\begin{gathered} 65=(1)/(2)*9.8* t^2 \\ t=\sqrt[]{(2*65)/(9.8)} \\ =3.64\text{ s} \end{gathered}`

Thus the time of flight of the diver is 3.64 s

The initial velocity of the diver is given by the equation,

`v=(R)/(t)`

On substituting the known values,

`\begin{gathered} v=(25)/(3.64) \\ =6.87\text{ m/s} \end{gathered}`

Thus the diver has to jump with an initial velocity of 6.87 m/s