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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 Î¸ 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 40âˆš2 m/s at 45Âº ****Find: ****(a) velocity and ****(b) displacement of the particle after 2 s. (g = 10 m/s ^{2})**

(b) 100 m

As horizontal acceleration would be zero.

v

s

A : position of particle at time = 0.

B : position of particle at time = t.

As vertical acceleration would be - g

v

= u sin Î¸ = gt

= 40 - 20

= 20 m/s

âˆ´

i.e.,

i.e.,

or

or

or t

i.e., t = 1 s and 3 s

u = 40 m/s, Î¸ = 60Â°

âˆ´ u

Thus, v

As Ï† = 45Â°

âˆ´ y

Thus,

= 20âˆš2 m/s

Before reaching highest point

v

âˆ´ 20 = 40 sin 60Â° - 10 t

or

â‡’

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 Î±**Sol: ****Ques 8: What is the change in velocity in the above question?****Ans:** 2u sin Î± (downwards)**Sol: **Change in velocity

= (- u sin Î±) - (+ u sin Î±)

= - 2u sin Î±

= 2u sin Î± (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 H_{max} 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 - 5t^{2} = 0**i.e., **

t = 0 belongs to the initial point of projection of the particle.

âˆ´

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 60Â° with horizontal with a speed v = 20 m/s. Taking g = 10 m/s ^{2}. Find the time after which the speed of the particle remains half of its initial speed.**

â‡’

= 10/v

= 10/10

[as, v = 20/2(given)]

= 1 i.e., Ï† = 0Â°

âˆ´ Speed will be half of its initial value at the highest point where Ï† = 0Â°.

Thus,

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