Question

A soldier jumps out from an aeroplane with a parachute .After dropping through a distance of 19.6 m he opens the parachute and decelarates at the rate of 1 m/s . If he reaches the ground with a speed of 4.6 m/s , how long was he in air

Answer 1

He was in air for 15 seconds
Like comment follow me on IG and brainly follow me on IG@hashtagger_owais_

Answer 2

Answer :

➙ The soldier was 17 sec in air.

Explanation :

Given :

➙ Distance = 19.6 m

➙ Decelerate = 1 m/s

➙ Speed = 4.6 m/s

To find :

➙ how long was he in air

Solution :

Using equation of motion

$\sf {\implies}\:s = ut+\dfrac{1}{2}\: gt^2$

$\sf {\implies}\:19.6=0+\dfrac{1}{2}\times9.8\times t^2$

$\sf {\implies}\:t=\sqrt{\dfrac{19.6}{4.9}}$

$\qquad\boxed {\sf {t = 2\ sec}}$

$\rule {300}{1.5}$

Using equation of motion

$\sf {\implies}\:v = u+gt$

$\sf {\implies}\:v = 0+9.8\times2$

$\qquad\boxed {\sf {v = 19.6\ m/s}}$

$\rule {300}{1.5}$

Decelerates at 1 m/s and reach the ground with the 4.6 m/s.

So, the final velocity is 4.6 m/s

The time taken :

$\sf {\implies}\:v = u-at'$

$\sf {\implies}\:4.6=19.6-1\times t'$

$\sf {\implies}\:t'=4.6-19.6$

$\qquad \boxed {\tt {t'=15\ s}}$

$\rule {300}{1.5}$

Where t' is the time taken to reach the ground

The Required Time is :

$\sf {\implies}\:T =t+t'$

$\sf {\implies}\:T = 2+15$

$\qquad\boxed {\sf {T= 17\ s}}$

$\rule {300}{1.5}$

$\therefore$The soldier was 17 sec in air.