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,

$h=\frac{1}{2}gt^2$

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

On substituting the known values,

$\begin{gathered} 65=\frac{1}{2}\times9.8\times t^2 \\ t=\sqrt[]{\frac{2\times65}{9.8}} \\ =3.64\text{ s} \end{gathered}$

Thus the time of flight of the diver is 3.64 s

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

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

On substituting the known values,

$\begin{gathered} v=\frac{25}{3.64} \\ =6.87\text{ m/s} \end{gathered}$

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