Question

The height of a helicopter above the ground is given by $h = 3.00{t^3}$ where h is in meters and $t$ is in seconds. At $t = 2.00\,s$ the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Answer 1

We are asked to find the time after which the mailbag reach the ground. We can start by writing down the given quantities. We can then move onto finding the initial height at the given time. Then in order to find the velocity, we can differentiate the formula given for height with respect to time. We can then find the time taken by setting the final distance as zero. This leads us to the required solution.

Formulas used:
The final value of displacement is given by,
${h_f} = {h_i} + {v_i}t + \dfrac{1}{2}a{t^2}$
Where $a$ is the acceleration, ${v_i}$ is the initial velocity, ${h_i}$ is the initial height, ${h_f}$ is the final height and $t$ be the time.

Complete step by step answer:
We can start by writing down the given values,
The height or distance is given by $h = 3.00{t^3}$.
The time is given as $t = 2.00s$.
We can find the distance or height at which the helicopter is at initially using the formula,
$h = 3.00{t^3}$
Substituting, we get
${h_i} = 3.00{t^3} = 3 \times 2 \times 2 \times 2 = 24m$

Now that we have the value of initial height, we can find the value of velocity by differentiating it with time

\Rightarrow {v_y} = 9.00{t^2} \\\ \Rightarrow {v_y} = 9 \times 2 \times 2 \\\ \Rightarrow {v_y} = 36\,m/s$$ We can find the value of final position of the mailbag using the formula, $${h_f} = {h_i} + {v_i}t + \dfrac{1}{2}a{t^2}$$ We set the value of final distance as zero and get, $$0 = {h_i} + {v_i}t + \dfrac{1}{2}a{t^2}$$ The time can be found by solving the equation, $$\dfrac{1}{2} \times 9.8 \times {t^2} + 36 \times t + 24 = 0$$ We can use the quadratic formula $$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$$ We get the value $$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\\ \Rightarrow x = \dfrac{{ - 36 \pm \sqrt {{{36}^2} - 4 \times 4.9 \times 24} }}{{2 \times 4.9}} \\\ \therefore x = 7.96\,s$$ **Hence,the mailbag reaches the ground in 7.96 s.** **Note:** We might get confused as to why we are using the initial value of distance or height to get time. This is because, if the helicopter releases the mailbag after a particular time, the free falling of the mailbag starts with upward velocity. The value of time can only be positive and hence we discard the negative value.