JEE Exam  >  JEE Notes  >  Motion under Gravity

Motion under Gravity - JEE PDF Download

2. Motion under gravity :

I format : (When a body is thrown vertically upward)

It includes two types of motion 
(i) Deaccelerated motion from A to B because the direction of velocity and acceleration is opposite. So speed 
decreases  
(ii) Accelerated motion from B to C because the direction of velocity and acceleration is same (downward). So speed increases 
Motion under Gravity - JEE
(a) Time of flight : 

It is the time taken by the particle to reach the ground. If the particle is thrown vertically upward with initial velocity u then
u= u
a = – g (take downward direction negative)
from equation
S = ut + Motion under Gravity - JEE  ⇒  Snet = 0 (when particle again reaches the ground)
t = T (time of flight)
   Motion under Gravity - JEE

(b) Maximum Height :  

from v2 = u2 + 2as

at maximum height v = 0, s = Hmax
⇒  0 = u2 –2 gHmax   ⇒   Motion under Gravity - JEE

(c) Final velocity

from v = u + at
v = vf a = – g   Motion under Gravity - JEE
v= – u
i.e. the body reaches the ground with the same speed with which it was thrown vertically upwards as it thrown vertically upward.

Question for Motion under Gravity
Try yourself:
Which type of motion occurs when a body is thrown vertically upward?
View Solution

(d) Time to reach half of the maximum height :-

ui = u   a = – g
from   
Motion under Gravity - JEE
H = 2ut – gt2   ⇒  gt– 2ut + H = 0

 Motion under Gravity - JEE

⇒  Motion under Gravity - JEEMotion under Gravity - JEE
Motion under Gravity - JEE     (.....1)

Equation 1 gives two value of time which corresponds to 
t1 =  Motion under Gravity - JEE(from ground to Hmax/2 in upward motion)
t2 = Motion under Gravity - JEE (from  ground to Hmax/2 in downward motion)

(e) Time to reach any general height h

Motion under Gravity - JEE
Let us assume that particle reaches from A to B in time t&  from A to C is time t2

Motion under Gravity - JEE
gt2 – 2ut + 2h = 0
Motion under Gravity - JEE
⇒   t1  + t2 = T (Time of flight)

II Format (Free fall) : 

A body released near the surface of the earth is accelerated downward under the influence of force of gravity.

(a) Time of Flight :

Motion under Gravity - JEE

Motion under Gravity - JEE
S = – H, u = 0, a = – g
t = T (Let assume)
⇒ – H = (0)T –  Motion under Gravity - JEE

(b) Final Velocity when body reaches the ground 

from v2 – u2 = 2as
s = – H   v = v u = 0 a = – g
⇒   – 0  = 2 (–g) (–H)  ⇒  vfMotion under Gravity - JEE

Ex.17 A ball is thrown vertically upwards with a velocity u from the ground. The ball attains a maximum height Hmax. Then find out the time and displacement at which ball have half of the maximum speed.

Sol. Maximum speed of the ball is u  At point B and C ball have speed u/2 but direction 
Motion under Gravity - JEE
is opposite so from
v = u + at
Let t1 is the time taken by the ball from point A to B and t2 is  the time taken by the ball from A to C
Motion under Gravity - JEE
Motion under Gravity - JEE
  Motion under Gravity - JEE

Ex.18 A ball thrown vertically upwards with a speed of 19.6 ms–1 from the top of a tower returns to the earth in 6 s. Find the height of the tower.

Sol. Here u = 19.6 ms–1
g = –9.8 ms–2
Net displacement, s = – h
Negative sign is taken because displacement is in  the opposite direction of initial velocity.

Motion under Gravity - JEE
Motion under Gravity - JEE
∴  – h = 19.6 × 6 + 1/2 × (–9.8) × 62
= 117.6 – 176.4 = –58.8
or h = 58.8 m

Ex.19 A ball is thrown vertically upwards with a velocity of 20 ms–1 from the top of a multi-storeyed building. The height of the point from where the ball is thrown is 25 m from the ground. (i) How high will the ball rise and (ii) how long will it be before the ball hits the ground?

Sol. (i) Here u = +20 ms–1, g = –10 ms–2 
At the highest point, v = 0
Suppose the ball rises to the height h from the point of projection.
As v2 – u2 = 2gs
   ∴   02 – 202 = 2 × (–10) × h    or h = + 20 m.
(ii) Net displacement, s = –25 m
Negative sign is taken because displacement is in the opposite direction of initial velocity.
As s = ut + 
 Motion under Gravity - JEE
or 5t2 – 20t – 25 = 0 or t2 – 4t – 5 = 0
or (t+ 1) (t – 5) = 0
As t ≠ –1,  so t = 5s.

Ex.20 A ball thrown up is caught by the thrower after 4s. How high did it go and with what velocity was it thrown ? How far was it below the highest point 3 s after it was thrown?

Sol. As time of ascent = time of descent 
∴ Time taken by the ball to reach the highest point = 2 s
For upward motion of the ball : u = ?,  v = 0, t = 2s, g = – 9.8 ms–2
As v = u + gt
∴  0 = u – 9.8 × 2 
or u = 19.6 ms–1
Maximum height attained by the ball is given by 
Motion under Gravity - JEE
Displacement of the ball in 3 s,
s = 19.6 × 3 + Motion under Gravity - JEE × (–9.8) × 32     = 58.8 – 44.1 = 14.7 m
Distance of the ball from the highest point 3 s after it was thrown  
= 19.6 – 14.7 = 4.9 m.

Ex.21 A balloon is ascending at the rate of 9.8 ms–1 at a height of 39.2 m above the ground when a food packet is dropped from the balloon. After how much time and with what velocity does it reach the ground? 
 Take g = 9.8 ms–2

Sol. Initially the food packet attains the upward velocity of the balloon, so
u = 9.8 ms–1, g = 9.8 ms–2 , s = –39.2 m
Here s is taken negative because it is in the opposite direction of initial velocity.
Using, s = ut + 1/2 gt2 , we get
– 39.2 = 9.8 t – 1/2× 9.8 t2
or  4.9 t2 – 9.8 t – 39.2 = 0    or t2 – 2t – 8 =0
or (t – 4) (t + 2) = 0      or t = 4s or – 2s
As time is never negative, so t = 4s.

Velocity with which the food packet reaches the ground is 
v = u + gt = 9.8 – 9.8 × 4 = – 29.4 ms–1.
Negative sign shows that the velocity is directed vertically downwards.
When a particle is dropped then it will automatically attains the velocity of the frame at that time.

Ex.22 Two balls are thrown simultaneously, A vertically upwards with a speed of 20 ms–1 from the ground, and B vertically downwards from a height of 40 m with the same speed and along the same line of motion. At what points do the two balls collide? Take g = 9.8 ms–2.

Sol. Suppose the two balls meet at a height of x from the ground after time t s from the start.
For upward motion of balls A :
u = 20 ms–1, g = – 9.8 ms–2
s = ut + 1/2gt2
x = 20 t – 1/2 × 9.8 t2 = 20t – 4.9 t2 ...(i)
For downward motion of ball B,

Motion under Gravity - JEE
40 – x = 20 × t + 1/2  × 9.8 t2
= 20t + 4.9 t2 ... (ii)
Adding (i) and (ii), 40 = 40 t   or t = 1 s
From (i), x = 20 × 1 – 4.9 × (1)2 = 15.1 m
Hence the two balls will collide after 1 s at a height of 15.1 m from the ground.

The document Motion under Gravity - JEE is a part of JEE category.
All you need of JEE at this link: JEE

Top Courses for JEE

FAQs on Motion under Gravity - JEE

1. What is motion under gravity?
Ans. Motion under gravity refers to the movement of objects under the influence of the Earth's gravitational force. It is also known as free-fall motion, where objects experience a constant acceleration due to gravity until they hit the ground or a surface.
2. What is the acceleration due to gravity?
Ans. The acceleration due to gravity is the acceleration experienced by objects due to the Earth's gravitational force. Its value is approximately 9.8 m/s^2 and is denoted by the symbol 'g'. This value varies slightly with the altitude and latitude of the object.
3. What is the difference between mass and weight?
Ans. Mass is the amount of matter present in an object and is measured in kilograms. Weight, on the other hand, is the force with which an object is pulled towards the Earth due to gravity and is measured in Newtons. The weight of an object depends on its mass and the acceleration due to gravity.
4. How does air resistance affect motion under gravity?
Ans. Air resistance is the force exerted by air molecules on objects moving through the air. It opposes the motion of the object and reduces its speed. In the case of motion under gravity, air resistance acts in the opposite direction to the gravitational force, and the net force acting on the object decreases, resulting in a slower rate of acceleration.
5. What are some real-world applications of motion under gravity?
Ans. Motion under gravity is used in many real-world applications, such as skydiving, bungee jumping, and projectile motion. It is also used in sports like long jump, high jump, and pole vaulting, where athletes use their knowledge of motion under gravity to maximize their performance. Additionally, motion under gravity is essential in understanding the behavior of objects in space, such as satellites and planets.
Download as PDF
Explore Courses for JEE exam

Top Courses for JEE

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

video lectures

,

Viva Questions

,

Summary

,

mock tests for examination

,

Objective type Questions

,

Important questions

,

Exam

,

ppt

,

Previous Year Questions with Solutions

,

Motion under Gravity - JEE

,

Extra Questions

,

practice quizzes

,

past year papers

,

Motion under Gravity - JEE

,

Sample Paper

,

study material

,

pdf

,

MCQs

,

Semester Notes

,

Motion under Gravity - JEE

,

shortcuts and tricks

,

Free

;