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DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE

Document Description: DC Pandey Solutions: Projectile Motion - 1 for JEE 2022 is part of DC Pandey Solutions for JEE Physics preparation. The notes and questions for DC Pandey Solutions: Projectile Motion - 1 have been prepared according to the JEE exam syllabus. Information about DC Pandey Solutions: Projectile Motion - 1 covers topics like Introductory Exercise 4.1 and DC Pandey Solutions: Projectile Motion - 1 Example, for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for DC Pandey Solutions: Projectile Motion - 1.

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Table of contents
Introductory Exercise 4.1
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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.

DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE

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 angleDC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE with horizontal, (b) 100 m

u = 40√2 m/s, θ = 45°
As horizontal acceleration would be zero.
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
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
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
i.e., DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
i.e., DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE

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 

DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
or  DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
or DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
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.DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE

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°
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
∴ yy = vx = 20 m/s
Thus, DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
= 20√2 m/s
Before reaching highest point
vy = uy + (- g) t

∴ 20 = 40 sin 60° - 10 t
or DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
After attaining highest point
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
i.e., DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
or  DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE

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

DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE

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

Change in velocity
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
= (- 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 DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE 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

DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
y-coordinate will be zero when 4t - 5t2 = 0
i.e., DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
t = 0 belongs to the initial point of projection of the particle.
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
i.e., x = 0 m
At t = 0.8 s,
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
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.DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE

vx = ux = 10 m/s

DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
= 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,  DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Projectile Motion - 1 Notes | Study DC Pandey Solutions for JEE Physics - JEE

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