Question

A helicopter is ascending vertically with a speed of 5.40 m/s. At a height of 105 m above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground? [Hint: What is v₀ for the package?]

Short answer

The obtained solution is $t=4.108\mathrm{s}$.

Step-by-step solution

Step 1. Kinematic equations 

Kinematics is one of the subfields of physics that describes the movement of a certain component without considering force.

Given data.

The speed of the helicopter is $v=5.40\mathrm{m}/\mathrm{s}$.

The height above the Earth is $h=105\mathrm{m}$.

The relation from the equation of motion is given by.

$\begin{array}{c}h=vt+\frac{1}{2}g{t}^2\\ t^2+2\frac{v}{g}t-\frac{2h}{g}=0\end{array}$

Here, g is the gravitational acceleration, and t is the time.

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

$\begin{array}{c}{t}^2+2\left(\frac{5.40\mathrm{m}/\mathrm{s}}{-9.81\mathrm{m}/{\mathrm{s}}^2}\right)t-\frac{2\left(-105\mathrm{m}\right)}{-9.81\mathrm{m}/{\mathrm{s}}^2}=0\\ t^2+1.1t-21.4=0\end{array}$

Step 2. Calculation of time

On solving the above equation, you get;

$\begin{array}{l}t=\frac{-1.1\pm \sqrt{{\left(1.1\right)}^2-4\left(1\right)\left(-21.4\right)}}{2\times 1}\\ t=4.108 \mathrm{s},-5.208 \mathrm{s}\end{array}$

Thus, $t=4.108 \mathrm{s}$is the required time of the helicopter.