Chapter 2, Problem 39P

To determine

The time taken by the package to reach the ground

Answer to Problem 39P

5.6 seconds.

Explanation of Solution

Given:

Velocity of the helicopter when package is dropped $=v_h=5.20\frac{\text{m}}{\text{s}}$

Initial velocity of the package as it dropped from the helicopter $=v_i=v_h=5.20\frac{\text{m}}{\text{s}}$

Time taken by package to reach the ground $=t$

Acceleration experienced by package = acceleration due to gravity $=a=-9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Height of the package when it is dropped $=h=125\text{m}$

Initial position of package as it leaves the helicopter $=y_i=h=125\text{m}$

Final position of package as it reach the ground $=y_f=0\text{m}$

Formula Used:

Kinematics equation that relates initial position, final position, initial velocity, acceleration and time of an object is given as

  $y_f=y_i+v_{i}t+(0.5)a{t}^2$

Calculation:

Consider the motion of package after it is dropped from the helicopter.

Substituting the values,

  $\begin{array}{l}0=125+(5.20)t+(0.5)(-9.8)t^2\\ 0=125+(5.20)t-(4.9)t^2\\ (4.9)t^2-(5.20)t-125=0\\ t=5.6\sec \end{array}$

Negative term is ignored.

Conclusion:

Therefore, time taken by the package to reach the ground is 5.6 seconds.