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

## DC Pandey (Questions & Solutions) of Physics: NEET

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## NEET : 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 NEET Course DC Pandey (Questions & Solutions) of Physics: NEET.
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Introductory Exercise 4.1

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

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

Ques 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:    Ques 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 angle with horizontal
(b) 100 m
Sol: u = 40√2 m/s, θ = 45°
As horizontal acceleration would be zero. 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    i.e.,  i.e., Ques 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
Sol: or or or  t2 = 4t + 3 = 0
i.e., t = 1  s and 3 s

Ques 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: Sol: 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° ∴ yy = vx = 20 m/s
Thus, = 20√2 m/s
Before reaching highest point
vy = uy + (- g) t

∴ 20 = 40 sin 60° - 10 t
or  After attaining highest point i.e., or Ques 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
Sol:   Ques 8: What is the change in velocity in the above question?
Ans: 2u sin α (downwards)
Sol: Change in velocity = (- u sin α) - (+ u sin α)
= - 2u sin α
= 2u sin α (downward)

Ques 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.
Sol: 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.

Ques 10: A body is projected up such that its position vector varies with time as 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
Sol: y-coordinate will be zero when 4t - 5t2 = 0
i.e., t = 0 belongs to the initial point of projection of the particle. i.e., x = 0 m
At t = 0.8 s, i.e.,   x = 2.4 m

Ques 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: Sol: vx = ux = 10 m/s   = 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,  Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

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