Question

a man jumps from a stationay helicopter at a height of 600m above the ground after falling first 100m, he opens the parachute . the parachute immediately reduces the velocity of man to half its value then becomes constant. find total time of motion​

Answer 1

Given:

• Parachute opened after 100 m
• Velocity became half and constant.

To find:

• Total time taken?

Calculation:

Time for falling the first 100 metres will be:

$h = ut + \dfrac{1}{2} g {t}^{2}$

• u will be zero.

$\implies 100 = 5 {t}^{2}$

$\implies {t}^{2} = 20$

$\implies t = 4.47 \: sec$

Now, velocity at end of 100 metres will be :

$v = u + at$

$\implies \: v = 0 + 10(4.47)$

$\implies \: v =44.7 \: m {s}^{ - 1}$

Now, parachute made the velocity half (i.e. 22.35 m/s)

The rest 600-100 = 500 m is travelled with this constant velocity.

So, rest time needed:

$t_{2} = \dfrac{500}{22.35} = 22.37 \: sec$

So, total time needed = 4.47 + 22.37 = 26.84 sec.

#SPJ3

Answer 2

Answer:

The total time of motion is $12\sqrt{5}$ sec.

Explanation:

Given height of helicopter from ground, $h=600m$

          distance travelled after falling from helicopter $s= 100 m$

Let this distance is covered in time $t_{1}$ seconds, with the acceleration due to gravity, $g=10m/s^{2}$

Using the equation of motion.

                           $s=ut+\frac{1}{2} gt^{2}$

as the initial velocity $u=0,$

                $\implies100=\frac{1}{2} \times 10\times t_{1} ^{2}\implies 100=5 t_{1} ^{2}$

           $\implies t_{1} ^{2}=20\implies t_{1} =2\sqrt{5}s$

Using equation of motion to find the velocity at time $t_{1}$,

                         $v^{2} -u^{2} =2gs$

                  $\implies v^{2} =2\times 10\times 100=2000$

                  $\implies v=20\sqrt{5}m/s$

When the parachute is opened, the velocity is halved.

Therefore, $v=\frac{20\sqrt{5}}{2} ={10\sqrt{5} m/s$

After this the velocity remains constant for remaining travel of distance i.e., $600-100=500m$

The time taken to travel this distance,

                           $t_{2} =\frac{d}{v}= \frac{500}{10\sqrt{5} }$

                               $=\frac{50\sqrt{5}}{\sqrt{5}\sqrt{5} }=10\sqrt{5}}}s$

Then the total time,  $t=t_{1} +t_{2}$

                                    $=2\sqrt{5}+10\sqrt{5}=12\sqrt{5} s$

Hence, the total time of motion is $12\sqrt{5}$ sec.