Question

Pochinki, a city in the Erangel map of PUBG, is one of the deadliest places to drop. A novice player, in lure of maximum loot aims to have a drop at Pochinki is released from a helicopter flying horizontally at a constant velocity of $40m{s^{ - 1}}$ . How far from the city of Pochinki (in $km$ ), horizontally, the player needs to drop if the helicopter is flying at a height of $180m$ from the ground. $\left[ {g = 10m/{s^2}} \right]$ .

Answer 1

Hint : To solve this question, we have to find out the time of flight of the dropped player. For this we have to use the second equation of motion in the vertical direction. Then, using this value of time, we can calculate the required horizontal distance.

Formula used: The formula used to solve this question is given by,
$s = ut + \dfrac{1}{2}a{t^2}$ , here $s$ is the displacement covered by a particle in time $t$ which moves with an initial velocity of $u$ and an acceleration of $a$ .

Complete step by step answer
Let the horizontal distance of the city of Pochinki from the point of drop of the player be $x$ .
Let $T$ be the time of flight of the player.
We first consider the vertical motion of the player. From the second kinematic equation of motion we have
$s = ut + \dfrac{1}{2}a{t^2}$ ........................(1)
As the player drops from the helicopter, so his initial velocity in the vertical direction is zero. That is, $u = 0$ . For the motion in the vertical direction, the acceleration is equal to the acceleration due to gravity, that is, $a = g$ . The vertical displacement covered by the player is the height of the helicopter, that is, $s = H$ in the time equal to the time of flight of the player, that is, $t = T$ . Substituting these in (1) we get
$H = 0T + \dfrac{1}{2}g{T^2}$
$\Rightarrow H = \dfrac{1}{2}g{T^2}$
According to the question, we have $H = 180m$ , and $g = 10m/{s^2}$ . Substituting these above, we get $180 = \dfrac{1}{2} \times 10 \times {T^2}$
$\Rightarrow {T^2} = 36$
Taking square root both the sides, we get
$T = 6s$
So the time of flight of the player is equal to $6s$ .
Now, as the player drops from the helicopter which is flying horizontally at a constant velocity of $40m/s$ , so the velocity of the player in the horizontal direction is $v = 40m/s$ which remains constant as no horizontal force acts on the player. So the horizontal displacement covered by the player is given by
$x = vT$
$\Rightarrow x = 40 \times 6 = 240m$
We know that $1m = {10^{ - 3}}km$ . So we get
$x = 240 \times {10^{ - 3}}km$
$\Rightarrow x = 0.24km$
Hence the player needs to drop at a horizontal distance of $0.24km$ .

Note
We should not forget to convert the final value of the horizontal distance, which is calculated in meters, into kilometers. To avoid this mistake, we can assume the distance $x$ to be in kilometers, and then we can substitute it in the formula after converting into meters.