Question

A parachutist bails out from an aeroplane and after dropping through a distance of $40m$ opens the parachute and decelerates at $2m{s^{ - 2}}$. He reaches the ground with a speed of $2m{s^{ - 1}}$, how long was he in the air? At what height did he bail out from the plane?

Answer 1

The given question is totally based on the application of Newton’s equations of motion. We need to apply the equations of motion in order to find the required values of time and height. After that we can finally conclude with the correct solution of the given question.

Complete step by step solution:
Since the parachutist bails out from the aeroplane, his initial velocity will be zero.
Now, we need to find the final velocity of the parachutist by using Newton’s third equation of motion. So, we can write it as,
$\Rightarrow {v^2} - {o^2} = 2 \times 9.8 \times 40$
$\Rightarrow {v^2} = 784$
$\therefore v = \sqrt {784} = 28m{s^{ - 1}}$
Now, we need to find the time taken to cover this distance.
We need to apply first equation of motion, i.e. $v = u + at$
Now, putting the values in above equation, we get,
$\Rightarrow 28 - 0 = 9.8{t_1}$
$\therefore {t_1} = \dfrac{{28}}{{9.8}} = 2.85s$
Now, he reached the ground with a velocity of $2m{s^{ - 1}}$ and deceleration of $2m{s^{ - 2}}$.
The initial velocity for this condition will be $28m{s^{ - 1}}$.
Let us assume that he travelled a distance $h$.
So, by applying third equation of motion we can write,
$\Rightarrow {2^2} - {28^2} = 2 \times ( - 2)h$
$\Rightarrow 4 - 784 = - 4h$
$\Rightarrow - 780 = - 4h$
$\therefore h = \dfrac{{780}}{4} = 195m$
Now the time taken to cover this distance can be found by using the first equation again. So, we can write it as,
$\Rightarrow 2 = 28 - 2{t_2}$
$\Rightarrow 2 - 28 = - 2{t_2}$
$\Rightarrow - 26 = - 2{t_2}$
$\therefore {t_2} = \dfrac{{26}}{2} = 13s$
Therefore, the total time taken when he was in air is given by,
$\Rightarrow t = {t_1} + {t_2} = 2.85 + 13 = 15.85s$
And the height at which he bail out is $195m + 40m = 235m$

Note: When a body falls freely from a height at that point only acceleration due to gravity acts on the body. The initial velocity at that point is zero. Deceleration is the negative acceleration and it is always denoted with a negative sign.