Chapter 2, Problem 66GP

To determine

The minimum height at which the fish must spot the pelican to escape

Answer to Problem 66GP

  $3.5\text{m}$

Explanation of Solution

Given:

For the motion of pelican from top to the surface of water:

Initial velocity of the pelican at the start of diver $=v_o=0\frac{\text{m}}{\text{s}}$

Height at which the pelican starts the dive $=h=16\text{m}$

Initial position of the pelican at the top at the start of dive $=y_o=h=16\text{m}$

Final position of the pelican at the surface of water $=y_f=0\text{m}$

Time taken by the pelican to reach the surface of water $=t$

Acceleration experienced by the pelican during free fall $=a=g=-9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Time taken by the fish to escape $=t_{esc}=0.2s$

For the motion of pelican from top to the height where fish spots it:

Time of travel for the pelican when it is spotted by the fish at the right time $=t^{\text{'}}=t-t_{esc}=t-0.2$

Initial velocity of the pelican at the start of diver $=v_o=0\frac{\text{m}}{\text{s}}$

Height at which the pelican gets spotted by fish $=H$

Final position of the pelican $=y_f^{\text{'}}=H$

Formula Used:

Position of an object at any time during constant accelerated motion is given as

  $y_f=y_o+v_{o}t+(0.5)a{t}^2$

Calculation:

Consider the motion of the pelican from top to the surface of water:

Position of the pelican when at the surface of water is given as

  $y_f=y_o+v_{o}t+(0.5)a{t}^2$

Inserting the values

  $\begin{array}{l}0=16+(0)t+(0.5)(-9.8)t^2\\ t=1.8\sec \end{array}$

Consider the motion of the pelican from top to the height where fish spots it:

Position of the pelican where fish spots it is given as

  $y_f^{\text{'}}=y_o+v_o{t}^{\text{'}}+(0.5)a{t}^{\text{'}}{}^2$

Substituting the values

  $\begin{array}{l}H=16+(0)(t-0.2)+(0.5)(-9.8){(}^t\\ H=16+(0)(1.8-0.2)+(0.5)(-9.8){(}^{1.8}\\ H=16+(0)(1.8-0.2)+(0.5)(-9.8){(}^{1.8}\\ H=3.45\text{m}\\ H\approx 3.5\text{m}\end{array}$

Conclusion:

Therefore, minimum height at which the fish must spot the pelican is 3.5 m.