Question

A parachutist bails out and freely falls 62 m. Then the parachute opens, and thereafter she decelerates at 2.5 m/s2. She reaches the ground with a speed of 2.8 m/s. (a) How long is the parachutist in the air

Answer 1

Answer:

16.4 s

Explanation:

For AB

initial velocity, u = 0

final velocity, v = ?

acceleration, a = 9.8 m/s^2

distance, s = 62 m

Let the time taken is t.

Use second equation of motion

$s=ut+\frac{1}{2}at^{2}$

$62=0+0.5\times9.8\times t^{2}$

t = 3.56 s

Use first equation of motion

v = u + at

v = 0 - 9.8 x 3.56

v = - 34.89 m/s

For BC

Let the time taken is t'

initial velocity, v = 34.89 m/s

final velocity, v' = 2.8 m/s

a = 2.5 m/s^2

use first equation of motion

v' = v + a t'

2.8 = 34.89 - 2.5 t'

t' = 12.84 s

So, the total time taken

T = t + t' = 3.56 + 12.84 = 16.4 s