Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Physics For JEE

JEE : Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

The document Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev is a part of the JEE Course Physics For JEE.
All you need of JEE at this link: JEE

Introductory Exercise 4.1

Q.1. Projectile motion is a 3-dimensional motion. Is this statement true or false.
Ans. False

A particle projected at any angle with horizontal will always move in a plane and thus projectile motion is a 2-dimensional motion. The statement is thus false.

Q.2. Projectile motion (at low speeds) is uniformly accelerated motion. Is this statement true or false.
Ans. True

At high speed the projectile may go to a place where acceleration due to gravity has some different value and as such the motion may not be uniform accelerated.
The statement is thus true.

Q.3. A particle is projected with speed u at angle θ with vertical. 
Find: 
(a) time of flight 
(b) maximum height 
(c) range 
(d) maximum range and corresponding value of 0.
Ans.

Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Q.4. A particle is projected from ground with velocity 40√2 m/s at 45º 
Find: 
(a) velocity and 
(b) displacement of the particle after 2 s. (g = 10 m/s2)
Ans. (a) 20√5 m/s at angleProjectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev with horizontal, (b) 100 m

u = 40√2 m/s, θ = 45°
As horizontal acceleration would be zero.
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
vx = ux = u cos θ = 40 m/s
sx = uxt = (u cos θ) t = 80 m
A : position of particle at time = 0.
B : position of particle at time = t.
As vertical acceleration would be - g
vy = uy - gt
= u sin θ = gt
= 40 - 20
= 20 m/s
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
i.e., Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
i.e., Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Q.5. A particle is projected from ground with velocity 20√2 m/s at 45°. At what time particle is at height 15 m from ground? (g = 10 m/s2)
Ans. 1 s and 3 s 

Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
or  Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
or Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
or  t2 = 4t + 3 = 0
i.e., t = 1  s and 3 s

Q.6. A particle is projected from ground with velocity 40 m/s at 60° with horizontal. Find speed of particle when its velocity is making 45° with horizontal. Also find the times (s) when it happens, (g = 10 m/s2)
Ans.Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

See figure to the answer to question no. 4.
u = 40 m/s, θ = 60°
∴ ux = 40 cos 60° = 20 m/s
Thus, vx = ux = 20 m/s
As φ = 45°
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
∴ yy = vx = 20 m/s
Thus, Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
= 20√2 m/s
Before reaching highest point
vy = uy + (- g) t

∴ 20 = 40 sin 60° - 10 t
or Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
After attaining highest point
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
i.e., Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
or  Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Q.7. What is the average velocity of a particle projected from the ground with speed u at an angle a with the horizontal over a time interval from beginning till it strikes the ground again?
Ans. u cos α

Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Q.8. What is the change in velocity in the above question?
Ans. 2u sin α (downwards)

Change in velocity
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
= (- u sin α) - (+ u sin α)
= - 2u sin α
= 2u sin α (downward)

Q.9. Under what conditions the formulae of range, time of flight, and maximum height can be applied directly in case of a projectile motion?
Ans. Between two points lying on the same horizontal line.

Formulae for R, T and Hmax will be same if the projection point and the point where the particle lands are same and lie on a horizontal line.

Q.10. A body is projected up such that its position vector varies with time as Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev Here, t is in seconds.

Find the time and x-coordinate of particle when its y-coordinate is zero.
Ans. time = 0, 0.8 s, x-coordinate = 0, 2.4 m

Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
y-coordinate will be zero when 4t - 5t2 = 0
i.e., Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
t = 0 belongs to the initial point of projection of the particle.
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
i.e., x = 0 m
At t = 0.8 s,
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
i.e.,   x = 2.4 m

Q.11. A particle is projected at an angle 60° with horizontal with a speed v = 20 m/s. Taking g = 10 m/s2. Find the time after which the speed of the particle remains half of its initial speed.
Ans.Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

vx = ux = 10 m/s

Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
= 10/v
= 10/10
[as, v = 20/2(given)]
= 1 i.e.,    φ = 0°
∴ Speed will be half of its initial value at the highest point where φ = 0°.
Thus,  Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

shortcuts and tricks

,

MCQs

,

Previous Year Questions with Solutions

,

past year papers

,

Projectile Motion (Part - 1) - Physics

,

Objective type Questions

,

Extra Questions

,

Projectile Motion (Part - 1) - Physics

,

Semester Notes

,

Solution by DC Pandey NEET Notes | EduRev

,

mock tests for examination

,

Solution by DC Pandey NEET Notes | EduRev

,

Important questions

,

Viva Questions

,

Solution by DC Pandey NEET Notes | EduRev

,

practice quizzes

,

video lectures

,

pdf

,

ppt

,

Sample Paper

,

Projectile Motion (Part - 1) - Physics

,

Exam

,

study material

,

Summary

,

Free

;