Question

a parachutist falls out from aeroplane and after dropping through a distance of 40m opens the parachute and decelerates at 2ms if he reaches thr ground with a speed of 2 ms how long is he in the air at what height did he fall out from the plane

Answer 1

Consider when the parachutist falls from the plane and opens the parachute after travelling through 40 m.

When the parachutist moves in the plane, his initial vertical velocity is zero as the plane moves horizontally parallel to the ground. Thus, $\displaystyle\sf {u=0\ m\ s^{-1}.}$

Since the parachutist travels through 40m before opening the parachute, $\displaystyle\sf {s=40\ m.}$

The acceleration here, $\displaystyle\sf {a=g=10\ m\ s^{-2}}$

Then the time taken to travel through this distance will be, by second kinematic equation,

$\displaystyle\longrightarrow\sf{t=\sqrt{\dfrac {2s}{a}}}$

$\displaystyle\longrightarrow\sf{t=\sqrt{\dfrac {2\times 40}{10}}}$

$\displaystyle\longrightarrow\sf{t=2.83\ s}$

And the velocity attained after travelling 40 m or when he opens the parachute is, by first kinematic equation,

$\displaystyle\longrightarrow\sf{v=at}$

$\displaystyle\longrightarrow\sf{v=10\times2.83}$

$\displaystyle\longrightarrow\sf{v=28.3\ m\ s^{-1}}$

Now consider when the parachutist opens the parachute and moves it with reaching the ground.

The acceleration here is given.

$\displaystyle\longrightarrow\sf{a=-2\ m\ s^{-2}}$

The initial velocity is found earlier.

$\displaystyle\longrightarrow\sf{u=28.3\ m\ s^{-1}}$

Given that the final velocity, $\displaystyle\sf {v=2\ m\ s^{-1}}$

Then the time taken for this motion is, by first kinematic equation,

$\displaystyle\longrightarrow\sf{t=\dfrac {v-u}{a}}$

$\displaystyle\longrightarrow\sf{t=\dfrac {2-28.3}{-2}}$

$\displaystyle\longrightarrow\sf{t=13.15\ s}$

And the distance travelled is, by third kinematic equation,

$\displaystyle\longrightarrow\sf{s=\dfrac {v^2-u^2}{2a}}$

$\displaystyle\longrightarrow\sf{s=\dfrac{2^2-28.3^2}{2\times-2}}$

$\displaystyle\longrightarrow\sf{s=199.2\ m}$

Hence the total time by which the parachutist remain in the air is,

$\displaystyle\longrightarrow\sf{t=(2.83+13.15)\ s}$

$\displaystyle\longrightarrow\sf{\underline {\underline {t=15.98\ s}}}$

And the height at which the parachutist falls from the plane is,

$\displaystyle\longrightarrow\sf{h=(40+199.2)\ m}$

$\displaystyle\longrightarrow\sf{\underline{\underline {h=239.2\ m}}}$