Question

An aeroplane moving horizontally at $20{\text{ }}m{s^{ - 1}}\;$ drops a bag. What is the displacement of the bag after 5 seconds ? Given ($g = 10\,m{s^{ - 2}}$).

Answer 1

to answer this question we first need to know that we will be using equations of motion. First we will have to find both horizontal and vertical displacement using equations of motion. We will find horizontal displacement by speed, distance and time formula. And for vertical displacement we will apply a second equation of motion.

Formula used:
$d = s \times t$
Where, $d$= distance, $s$= speed and $t$= time.
$d = ut + \dfrac{1}{2}a{t^2}$
Where $d$- distance, $u$-initial velocity, $a$-acceleration and $t$ is time.

Complete step by step answer:
We will consider only horizontal motion now, we have been given the Initial horizontal velocity $20\,m{s^{ - 1}}$. Time of flight =$5s$. We know that distance is speed multiplied by time. So, the horizontal displacement is given by:
$d = s \times t$
$\Rightarrow d = 5 \times 20$
$\Rightarrow {d_h} = 100m$.

Now we will have to calculate vertical displacement also.
By second equation of motion we know that,
$d = ut + \dfrac{1}{2}a{t^2}$
Here ‘u’ is the initial velocity which is zero in the vertical case. And acceleration will be equal to acceleration due to gravity(g).
Vertical displacement is: $\dfrac{1}{2}g{t^2}$
$d = \dfrac{1}{2} \times 10 \times {5^2}$
$\Rightarrow d = \dfrac{1}{2} \times 10 \times 25$
$\Rightarrow d = 5 \times 25$
$\Rightarrow {d_v} = 125\,m$

Now we will apply the vector equation to solve this question further. We have found 2 components now, vertical as well as horizontal. Hence we need to find the resultant displacement, it can be found as:
$d = \sqrt {d_h^2 + d_v^2}$
$\Rightarrow d = \sqrt {{{100}^2} + {{125}^2}}$
$\Rightarrow d = \sqrt {10000 + 15625}$
$\Rightarrow d = \sqrt {25625}$
$\therefore d = 160.07\,m$

Hence the displacement of the bag after 5 seconds is 160.7 meters. Or approximately 160 m.

Note: While finding the vertical displacement of the bag, students might make a silly mistake by applying initial velocity as $20{\text{ }}m{s^{ - 1}}\;$. This will be wrong, you cannot apply this because it has been clearly mentioned in the question that its horizontal velocity is $20{\text{ }}m{s^{ - 1}}\;$ and not vertical.