Pelicans tuck their wings and free-fall straight down while diving for fish. Suppose a pelican starts its dive from a height of 14.0 m and cannot change its path once committed. If it takes a fish 0.20 s to perform an evasive action, at what minimum height must it spot the pelican to escape? Assume that the fish is at the surface of the water.

You can use the kinematics equation to study the motion of free-falling objects because these objects have uniform acceleration during their motion. If the initial velocity of an object of mass ‘m’ is ‘u’ and it covers a distance ‘s’ with the uniform acceleration ‘a’ during its motion, the kinematics equation of motion connecting these variables can be written as

$s=ut+\frac{1}{2}a{t}^2.$

The distance covered by the pelican to reach the surface of the water is s = 14.0 m.

The time taken by the fish to perform an evasive action is $t\text{'}=0.20\mathrm{s}$.

As the pelican falls freely in the downward direction, the acceleration of the pelican will be equal to the acceleration due to gravity, i.e., $a=g=-9.8\mathrm{m}/{\mathrm{s}}^2$(as the downward direction is considered negative).

The initial velocity of the pelican before starting its dive is $u=0\mathrm{m}/\mathrm{s}$.

Let the time taken by the pelican to reach the surface of the water be t.

Using the third equation of motion,

$\begin{array}{c}s=ut+\frac{1}{2}a{t}^2\\ -14.0\mathrm{m}=\left(0\mathrm{m}/\mathrm{s}\right)t-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)t^2\\ t^2=\frac{2\times \left(14.0\mathrm{m}\right)}{9.8\mathrm{m}/{\mathrm{s}}^2}\\ t^2=2.86{\mathrm{s}}^2\\ t=1.69\mathrm{s}\end{array}$

Consider that the fish would be able to escape if it spots the pelican at $t\text{'}$time from the start of the free fall of the pelican. So,

$\begin{array}{c}t\text{'}=t-t_1\\ =1.69-0.20\\ =1.49\mathrm{s}\end{array}$

Consider that the pelican will travel the distance $s\text{'}$in time $t\text{'}$.

Using the equation of motion,

$\begin{array}{c}s=ut\text{'}+\frac{1}{2}a{\left(t\text{'}\right)}^2\\ s=\left(0\mathrm{m}/\mathrm{s}\right)\left(1.49\mathrm{s}\right)-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(1.49\mathrm{s}\right)}^2\\ s=-10.9\mathrm{m}\end{array}$

The negative sign of the displacement shows that this is measured from top to bottom. The pelican’s motion is in the downward direction, and the downward direction is considered negative in the problem.

Thus, the minimum height from the surface of the water at which the fish should spot the pelican to escape is

$\begin{array}{c}s\text{'}-s=\left(14.0-10.9\right)\mathrm{m}\\ =3.1\mathrm{m}\end{array}$