#### Why Kaysons ?

#### Video lectures

Access over 500+ hours of video lectures 24*7, covering complete syllabus for JEE preparation.

#### Online Support

Practice over 30000+ questions starting from basic level to JEE advance level.

#### Live Doubt Clearing Session

Ask your doubts live everyday Join our live doubt clearing session conducted by our experts.

#### National Mock Tests

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.

#### Organized Learning

Proper planning to complete syllabus is the key to get a decent rank in JEE.

#### Test Series/Daily assignments

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.

#### SPEAK TO COUNSELLOR ? CLICK HERE

## 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. .

2010 | 2011 | 2012 | 2013 | 2014 | 2015 | 2016 | 2017 | 2018 | 2019 | 2020 | |

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 |

- 1
- 2
- 3
- 4

**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