Question

A parachutist bails out from an airplane and after dropping through a distance of 40 m, he opens the parachute and decelerates at $2\text{ m/}{\text{s}}^2$. If he reaches the ground at a speed of 2 m/s. (1) How long is he in the air? (2) At what height did he bail out from the plane?

Answer 1

Hint: The motion of the parachutist can be broken into two parts, first the free fall from the plane and second the accelerated motion after opening the parachute till reaching the ground.

Complete step by step answer:
We know that when a body is dropped from a certain height, it executes free fall with an acceleration equal to acceleration due to gravity. So when the parachutist bails out from the plane he undergoes free fall for a distance of 40 m. So the velocity he gains while falling 40 m is given by,
${\text{v}}^{\text{2}}=\text{2gs}$,
Where, v is the final velocity, s is 40 m and g is $\text{9}\text{.8m/}{\text{s}}^{\text{2}}$. So substituting this in equation we get,
${\text{v}}^{\text{2}}=784$
$\therefore \text{v}=28\text{ m/s}$ …… equation (1)
The time taken by the parachutist to fall a distance of 40 m is given by,
$\text{t}=\sqrt{{\displaystyle \frac{\text{2s}}{\text{g}}}}$
Substituting the values of s=40 m and g=9.8 and taking the root of the result, we get the time taken (t) as,
$\text{t}=2.857\approx 2.86\text{ seconds}$ …… equation (2)
This is for the first part of the motion, when the parachutist is in freefall.
For the second part of motion, the parachutist opens his parachute and decelerates at a rate of $2\text{ m/}{\text{s}}^2$. The final velocity when he reaches the ground is 2 m/s. The initial velocity of the parachutist at this point is 28 m/s which is known to us from equation (1).
So the time taken for the parachutist to reach the ground is given by the equation,
$\text{v}=\text{u}+\text{(-a)t}$
Where, v is the final velocity, u is the initial velocity and (-a) is the deceleration and t is the time taken. In our problem v=2 m/s, u= 28 m/s, a=2 $m/s^2$. So substituting in the equation we get
$\text{t}={\displaystyle \frac{\text{u-v}}{\text{a}}}={\displaystyle \frac{28-2}{2}}$
$\therefore \text{ t=13 seconds}$…. Equation (3)
The height through which the parachutist fall after opening the parachute can be calculated using the formula,
${\text{v}}^{\text{2}}-{\text{u}}^{\text{2}}=\text{2as}$
Where, v is the final velocity, u is the initial velocity, a is the deceleration and s is the distance travelled. Substituting the values of u, v and a we get,
$\text{s}={\displaystyle \frac{{(2)}^2-{(28)}^2}{(-4)}}={\displaystyle \frac{780}{4}}$
$\therefore \text{ s}=195\text{ m}$…..equation (4)
From equation (2) and (3), the total time taken by the parachutist to reach the ground is,
$\text{t}={\text{t}}_{\text{1}}+{\text{t}}_{\text{2}}$
$t=2.86+13$
$\therefore \text{ t}=15.86\text{ seconds}$
This is the answer for question 1.
The total distance travelled by the parachutist is,
$\text{s}={\text{s}}_{\text{1}}+{\text{s}}_{\text{2}}=40+195$
$\therefore \text{ s=235 m}$
This is the answer for the second question.

Note: If a body is under freefall in real life, the body will experience drag forces due to the air present.