Question

A parachutist jumps from an airplane moving with a velocity of u.parachutist opens and accelerates downwards with 2 m/$s^2$. He reaches the ground with a velocity of 4 m/s. How long does the parachutist remain in the air?

Answer 1

Step 1, Given data

The initial velocity of the parachute = 0

Acceleration = 2 m/$s^2$

Final velocity = 4 m/s

Step 2, Finding time

From the equation of motion

We know,

$v=u+at$

Putting all the values

$4=0+2\times t$

$Or,t=2sec$

Hence parachute remained in the sky for 2 sec.