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 decelerates 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

For calculation see the attachment.

total time he was in the air is 17 sec

Answer 2

Answer:

The soldier was 17 sec in air.

Explanation:

Given that,

Distance = 19.6 m

Decelerate = 1 m/s

Speed = 4.6 m/s

The distance covers 19.6 m in time t with the acceleration due to gravity.

Using equation of motion

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

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

$t=\sqrt{\dfrac{19.6}{4.9}}$

$t = 2\ sec$

The velocity after 2 sec

Using equation of motion

$v = u+gt$

$v = 0+9.8\times2$

$v = 19.6\ m/s$

When the parachutist open the parachute then the initial velocity is 19.6 m/s.

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 is given by

$v = u-at'$

$4.6=19.6-1\times t'$

$t'=4.6-19.6$

$t'=15\ s$

Where t' is the time taken to reach the ground

The total time is

$T =t+t'$

$T = 2+15$

$T= 17\ s$

Hence, The soldier was 17 sec in air.