A parachutist after bailing out falls 52.0 m without friction. When the parachute opens, she decelerates at 2.10 m/s² and reaches the ground with a speed of 2.90 m/s. (a) How long is the parachutist in the air? (b) At what height did the fall begin?

(a) Given that the parachute travels the distance say y₁ pf 52.0 m without friction, the time t₁ taken to travel that distance under action of gravity is given by equation (1) as:

Substitute the given values,

Therefore the parachutist took 3.25 s to fall by 52.0 .

Substitute the given values,

The negative sign in acceleration accounts for the fact that the parachutist decelerates.
Therefore the time t for which the parachutist is in the air is given as:

Round off to three significant figures,
t = 17.1 s
Therefore the parachutist was in the air for 17.1 s.
(b) The distance y2 travelled by the parachutist under the constant deceleration a is given using equation (3) and is derived as:
From the equation of kinematics, we have


Substitute the value in first equation

The speed (v) is given as the product of time t1 and the free fall acceleration, therefore by substituting the same, equation (3) is derived as:

The height y from which the fall began is given as:
y = y₁ + y₂
Substitute the given values of y₁ to be 52.0 m and the calculated value of y₂ in the equation above,
y = 52.0 m + 240.01 m
= 292.01 m
Round off to three significant figures,
y = 292 m
Therefore the parachutist fell from 292 m above the ground.