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Projectile Motion JEE( 2 D Motion>) Chapterwise Analysis, Weightage and Overview

With a total of 10 questions asked, however, this chapter has not seen many direct questions in the recent years. It is an interesting chapter as the students can figure out how to decompose 2D motions into respective axes. The basics should be covered for sure alongside the fundamental equations. .

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Projectile Motion

1 1 1 2 0 0 0 0 0 2 2

 

List of important topics in Projectile Motion JEE

  • Total Time of Flight
  • Horizontal Range
  • Maximum Height
  • Equation of Trajectory

Projectile Motion JEE Syllabus

  • Projectile Motion.
  • Motion in vertical direction
  • co-ordinates and velocity components of projectile
  • Projectile Motion In Inclined Plane,
  • Up the plane
  • Time of Flight
  • Range
  • Down The Plane.

Download Projectile Motion JEE Notes

 
Projectile Motion Download

Projectile Motion JEE Test Series 2022,2023,2024

List of FREE Tests  
Paper set 1 Download Questions Download Solutions

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Q.

As shown in the figure there is a particle of mass kg, is projected with speed 10 m/s at an angle  with horizontal  then match the following :

                                                     

          Column 1

           Column 2

(A)

Average velocity (in m/s) during half of the time of flight is

(p)

 

(B)

The time (in sec) after which the angle between velocity vector and acceleration vector becomes  

(q)

 

(C)

Horizontal range (m)

(r)

 

(D)

Change in linear momentum (N-s) when particle is at highest point

(s)

At an angle of  from horizontal

 

Q.

A particle is projected from a horizontal surface with velocity u at an angle  with the horizontal. The particle strikes the same horizontal surface after timeT at a distance R from point of projection. t represent the time at the which the velocity of the particle becomes perpendicular to the initial velocity vector during the flight. h represents the height of the particle at that instant. Hrepresents the maximum height of the particle.

          Column 1

                    Column 2

 (A)

  

 (p)

 t = Th = 0

 (B)

  

 (q)

  

 (C)

  

 (r)

  

 (D)

  

 (s)

 t = not possible, h = not possible

 

Q.

A ball is thrown at an angle  with the horizontal at a speed of 20 m/s towards a high wall at a distance d. If the ball strikes the wall, its horizontal velocity component reverses the direction without change in magnitude and the vertical velocity component remains same. Ball stops after hitting the ground. Match the statement of Column-I with the distance of the wall from the point of throw in column-II.  

          Column 1

           Column 2

 (A)

 Ball can strike the wall for

 (p)

 d = 8 m

 

 (B)

 Ball strikes the ground at a distance of 12  m from the wall

 (q)

 d = 10 m

 (C)

 Ball strikes the ground at a distance of 10  m from the wall

 (r)

 d = 15 m

 (D)

 Ball strikes  the ground at a distance of 5  m from the wall

 (s)

 d = 25 m

 

Q.

As shown in the figure there is a particle of mass kg, is projected with speed 10 m/s at an angle  with horizontal  then match the following :

                                                 

                 Column 1

                Column 2

 (A) 

 Average velocity (in m/s)  during half of the time of  flight is

 (p)

  

 (B)

 The time (in sec) after  which the angle between  velocity vector and  acceleration vector  becomes  

 (q)

     

 (C)

 Horizontal range (m)

 (r)

  

 (D)

 Change in linear  momentum (N-s) when  particle is at highest point

 (s)

 At an angle of  from horizontal

 

Q. What is the formula for Hmax and Rmax.?

u^2 / 2g

2UxUy /g

Q. A projectile(A) is thrown horizontally. Another projectile(B) is thrown at 40° with 30 m/s velocity. If same time is taken by the two projectile to reach the ground then Va = Vbcos theta. How? I have determined the time required by the two projectiles. Then tried to form a relation between x and v. But not being able to reach the answer?

they will cover equal displacement in y direction and time is also same to reach the groud .

therefore

for projectile A, y = 0*t + 1/2*g*t^2

for projectile B , y = -Vbsintheta *t - 1/2 g*t2

from equating these two here you will get the value of t