Question

A parachutist drops first freely from an aeroplane for 10s and then his parachute opens out. Now he descends with a net retardation of $2.5m{s^{ - 1}}$. If he bails out of the plane at a height of 2495m and $g = 10m{s^{ - 2}}$, his velocity on reaching the ground will be
(A) 5m/s
(B) 10m/s
(C) 15m/s
(D) 20m/s

Answer 1

Hint : We need to find the velocity of the parachutist after the free fall period. Use this velocity as initial velocity for retardation period.
Formula used: In this solution we will be using the following formulae;
$v = u + at$ where $v$ is the final velocity of the body accelerating uniformly, $u$ is the initial velocity, $a$ is the value of the acceleration and $t$ is the time elapsed.
$s = ut + \dfrac{1}{2}a{t^2}$ where $s$ is the distance covered by an accelerating body.
${v^2} = {u^2} + 2as$ where all variables are the same as defined above.

Complete Step-by-Step solution:
During the period where the parachutist was in free fall, he accelerates downward with an acceleration due to gravity. We can calculate the velocity of the parachutist after the 10 sec of free fall using
$v = u + at$ where $v$ is the final velocity of the body accelerating uniformly, $u$ is the initial velocity, $a$ is the value of the acceleration and $t$ is the time elapsed.
Hence,
$v = 0 + \left( {10} \right)\left( {10} \right) = 100m{s^{ - 1}}$ (since $a = g = 10m{s^{ - 2}}$)
The distance covered within this same time period can be given using
$s = ut + \dfrac{1}{2}a{t^2}$ where $s$ is the distance covered by an accelerating body.
Hence,
$s = 0 + \dfrac{1}{2}\left( {10} \right){\left( {10} \right)^2} = 500m$
Hence, the distance to the ground after the free fall, is
${s_r} = s - 500 = 2495 - 500 = 1995m$
Now for the velocity on reaching the ground we can use the formula
${v^2} = {u^2} + 2as$
Hence by inserting known values, we have
${v^2} = {\left( {100} \right)^2} + 2\left( { - 2.5} \right)\left( {1995} \right)$
By computation, we have
$\Rightarrow v = 5m/s$

Hence, the correct option is A

Note: For clarity, the acceleration in the equation is negative because after the parachute was opened, the parachutist was in retardation, which is negative acceleration. Sometimes, the equation of motion can be written as
${v^2} = {u^2} \pm 2as$ where the $+$ is used for acceleration, and $-$ is used for deceleration or retardation. In this case, the $a$ will always possess a positive value.