The cliff divers of Acapulco push off horizontally from rock platforms about 35 m above the water, but they must clear rocky outcrops at water level that extend out into the water 5.0 m from the base of the cliff directly under their launch point. See Fig. 3–53. What minimum push-off speed is necessary to clear the rocks? How long are they in the air?

Consider the downward direction as positive y and the origin as the point from which the divers push off the cliff. Also, in the vertical direction, the value of the initial velocity is equivalent to zero.

Given data:

The distance above the water is $d=35\mathrm{m}$.

The distance from the base of the cliff is $L=5\mathrm{m}$.

The formula to calculate the time of flight can be written as:

$d=ut+\frac{1}{2}g{t}^2$

Here, u is the initial velocity in the vertical direction whose value is zero, and g is the gravitational acceleration.

On plugging the values in the above relation, you get:

$\begin{array}{c}35\mathrm{m}=\left(0\right)t+\frac{1}{2}\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)t^2\\ t=2.67\mathrm{s}\end{array}$

Thus, $t=2.67\mathrm{s}$is the required time.

The formula to calculate speed can be written as:

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

On plugging the values in the above relation, you get:

$\begin{array}{l}v=\left(\frac{5\mathrm{m}}{2.67\mathrm{s}}\right)\\ v=1.87\mathrm{m}/\mathrm{s}\end{array}$

Thus, $v=1.87\mathrm{m}/\mathrm{s}$is the minimum push-off speed.