Question

A bag is dropped from a helicopter rising vertically at a constant speed of 2 m/s. The distance between the two after 2s is (Given $g = 9 \cdot 8m{s^{ - 2}}$).
A) 4.9 m.
B) 19.6 m.
C) 29.4 m.
D) 39.2 m.

Answer 1

The speed of the helicopter is in upwards and therefore the bag will initially move upwards to reach at a point where the final velocity becomes zero and then the bag will start falling down towards ground as the acceleration due to gravity is always in the downwards direction.

Formula used: The formula of first relation of Newton’s law of motion is given by,
$\Rightarrow v = u + at$
Where final velocity is v initial velocity is u the acceleration is a and the time taken is t.
The formula of the second relation of Newton’s law of motion is given by,
$\Rightarrow s = ut + \dfrac{1}{2}at$
Where displacement is s the initial velocity is u the acceleration is a and the time taken is t.

Complete step by step solution:
It is given in the problem that a bag is dropped from a helicopter rising vertically at a constant speed of $2\dfrac{m}{s}$ and we need to find the distance between the bag and the helicopter after 2s.
The formula of first relation of Newton’s law of motion is given by,
$\Rightarrow v = u + at$
Where final velocity is v initial velocity is u the acceleration is a and the time taken is t.
The final velocity of the bag is equal to,
$\Rightarrow v = u + at$
The initial velocity of the bag is zero and the time taken is equal to 2s.
$\Rightarrow v = 0 + \left( {9 \cdot 8} \right) \cdot 2$
$\Rightarrow v = 19 \cdot 6\dfrac{m}{s}$
The formula of the second relation of Newton’s law of motion is given by,
$\Rightarrow s = ut + \dfrac{1}{2}at$
Where displacement is s the initial velocity is u the acceleration is a and the time taken is t.
The distance between bag and the helicopter is equal to,
$\Rightarrow s = \left( 0 \right) \cdot t + \dfrac{1}{2} \times \left( {9 \cdot 8} \right) \times {\left( 2 \right)^2}$
$\Rightarrow s = \dfrac{1}{2} \times \left( {9 \cdot 8} \right) \times {\left( 2 \right)^2}$
$\Rightarrow s = 19 \cdot 6m$
The distance between the helicopter and the bag after 2 sec is equal to $s = 19 \cdot 6m$.

The correct answer for this problem is option (B).

Note: It is advisable to the students to understand and remember the relations in the formula of Newton’s law of motion. The initial velocity of the bag with respect to the helicopter is zero and as the time passes due to the acceleration due to gravity the velocity of the bag keeps on increasing till the bag reaches ground.