• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


Fundamentals Of Physics
Found in: Page 39

Answers without the blur.

Just sign up for free and you're in.


Short Answer

A parachutist bails out and freely falls 50 m. Then the parachute opens, and there after she decelerates at 2.0 m/s2 She reaches the ground with a speed of 3.0 m/s. (a) How long is the parachutist in the air? (b) At what height does the fall begin?

(a) The parachutist is in the air for 17 s .

(b) From 290 m the fall begins.

See the step by step solution

Step by Step Solution

Step 1: Given information

Δy=50 m

vf=3.0 m/s

Step 2: To understand the concept

The problem deals with the kinematic equations of motion. Kinematics is the studies of how a system of bodies moves without taking into account the forces or potential fields that influence the motion. The equations which are used in the study are known as kinematic equations of motion.


The displacement is given by,

Δy=v0t+12 at2

The final velocity is expressed as,

vf2 = v02+2a Δy

vf = v0+at

Step 3 (a): Calculation for time for which parachutist was in the air

During free fall the initial velocity parachutist is 0 m/s and acceleration of the is

a=g= -9.8 m/s2

According to the newton’s second kinematic equation,

Δy=v0t+12 at2

50 =0 - 0.5 9.8 t2

role="math" localid="1662968239356" t = 3.19 s 3.2 s

The final velocity of the parachutist in free fall can be calculated as below,

According to the newton’s third kinematic equation,

vf2 = v02+2a Δy

vf2 =2 9.850

vf = 31.3 m/s

This final velocity for free fall of the parachutist is her initial velocity after parachute is open.

According to the newton’s first kinematic equation,

vf = v0+at

3 = 31.3 - 2t

role="math" localid="1662968456612" t =31.3 -32

t=14.15 14 sec

Therefore, the time for which the parachutist is in the air is,

3.19+14.15=17.34 17 sec

Step 4 (b): Calculation of height at which the falling started

The displacement of the parachutist after parachute opened can be calculated as below,

According to the newton’s second kinematic equation,

Δy=v0t+12 at2


Δy= 242 240 m

Therefore, the height from which the fall begins is,

h=50+240 290 m

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.