Question

A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height h. Then find the time taken by the packet to reach the ground. [g is the acceleration due to gravity]
A. \[3.4\sqrt {\dfrac{h}{g}} \]
B. \[\sqrt {\dfrac{{2h}}{{3g}}} \]
C. \[\dfrac{2}{3}\sqrt {\dfrac{h}{g}} \]
D. \[1.8\sqrt {\dfrac{h}{g}} \]

Answer 1

Hint:Acceleration due to gravity is defined as the acceleration gained by an object due to gravitational force and its S.I. unit is \[m{s^{ - 2}}\]. It is a vector quantity where it has both magnitude and direction.

Formula Used:
From the equation of motion we have,
\[{v^2} = {u^2} + 2as\]
Where, v is the final velocity, u is the initial velocity, a is acceleration and s is displacement.

Complete step by step solution:
Consider a helicopter holding packet A, after some time it reaches a height h from the ground and drops the object. Now the packet will have some upward velocity v and it will go to a certain height and it will drop to its original position. Here we need to find the time taken by the packet to reach the ground. Before finding the time let us calculate the velocity v. since this helicopter is going upwards with an acceleration g.

Now consider the equation of motion we have,
\[{v^2} = {u^2} + 2as\]
This can be written as,
\[{v^2} - {u^2} = 2ah\]
Here, displacement s is replaced by h and acceleration is directed upwards therefore a is replaced by g. Initially, the helicopter was at rest, so the initial velocity u will be zero. Then,
\[{v^2} - 0 = 2gh\]
\[\Rightarrow {v^2} = 2gh\]
\[\Rightarrow v = \sqrt {2gh} \]
When this pocket is dropped, the moment it detaches from the helicopter, the acceleration of it would be acting downwards which is of value -g.

Now, let us apply the second equation of motion,
\[S = ut + \dfrac{1}{2}a{t^2}\]
Here, \[S = - g\] and now the initial velocity is, \[u = \sqrt {2gh} \]
Substitute the value in the above equation we get,
\[ - g = \sqrt {2gh} \times t + \dfrac{1}{2}g{t^2}\]
\[\dfrac{1}{2}g{t^2} - \sqrt {2gh} \times t - h = 0\]
By applying the quadratic formula, we get,
\[t = \dfrac{{\sqrt {2gh} \pm \sqrt {{{\left( { - \sqrt {2gh} } \right)}^2} - 4 \times \left( {\dfrac{1}{2}g} \right)\left( { - h} \right)} }}{g}\]
\[\Rightarrow t = \dfrac{{\sqrt {2gh} \pm \sqrt {\left( {2gh} \right) + \left( {2gh} \right)} }}{g}\]
\[\Rightarrow t = \dfrac{{\sqrt {2gh} \pm \sqrt {4gh} }}{g}\]
\[\Rightarrow t = \dfrac{{\sqrt {2gh} \pm 2\sqrt {gh} }}{g}\]
\[\Rightarrow t = \sqrt {gh} \left( {\dfrac{{\sqrt 2 \pm 2}}{g}} \right)\]

Here, we have two roots that are,
\[t = \sqrt {gh} \left( {\dfrac{{\sqrt 2 + 2}}{g}} \right)\]and \[t = \sqrt {gh} \left( {\dfrac{{\sqrt 2 - 2}}{g}} \right)\]
Since the second value was found to be negative, we consider the first value, then
\[t = \sqrt {\dfrac{h}{g}} \left( {\sqrt 2 + 2} \right)\]
\[\Rightarrow t = \sqrt {\dfrac{h}{g}} \left( {1.414 + 2} \right)\]
\[\therefore t = 3.41\sqrt {\dfrac{h}{g}} \]
Therefore, the time taken by the packet to reach the ground is \[3.41\sqrt {\dfrac{h}{g}}\].

Hence, option A is the correct answer.

Note: Here it is important to remember the sign of acceleration due to gravity. If the direction of motion of the object is opposite to the direction in which the gravitational force acts, then we consider the sign of the acceleration due to gravity as negative.