Question

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s². He reaches the ground with a speed of 3 m/s. At what height, did he bail out ? (Given g=9.8 m/s2 approximately)
A) 293
B) 111
C) 91
D) 182​

Answer 1

$\Large\bold{\underline{\underline{Answer:-}}}$

$\bf\huge\boxed{293 m}$

$\rule{200}{4}$

$\bf\small\bold{\underline{\underline{Step-By-Step\:Explanation:-}}}$

The speed of the person just before the parachute opens is:

$u = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 50}$

$u = \sqrt{980}$

Now

When the parachute opens and descends,

BY NEWTON'S LAW

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

putting the value

$9 - 980 = - 2 \times 2 \times h$

Here a is negative as perticles decelerates

$h = \frac{971}{4}$

$h = 242.75m$

Therefore, The total height of fall is

⇒242.75 + 50 = 292.45 m

=293 m (approximately )

Option —(A)

Answer 2

Answer:-

Given:-

He falls 50m without friction.

After opening he decelerates at the rate of 2m/s^2.

Reaches the ground with velocity = 3m/s.

Explanation:-

As he is under free fall the final velocity will be zero.

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

v = 0.

$u = \sqrt{2gh}$

$u = \sqrt{2 \times 9.8 \times 50}$

$u = \sqrt{980}$

or

$u = 31.3$ m/s

Now,the parachute is decelerated.

From third equation of motion.

${v}^{2} - {u}^{2} = 2as \:$

v = 3m/s.

$9 - 980 = 2 \times - 2 \times h$

$- 971 = - 4 \times h$

$h = \frac{971}{4}$

$h = 242.75 \: meters$

Now,

The total height of fall is

H = 242.75 + 50 = 292.75 meters.

H = 293 meters (approximately).

$\boxed{\boxed{H = 293 meters}}$

So,the person bailed out at 293 meters.