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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.**

**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/s ^{2})**

u = 40√2 m/s, θ = 45°

As horizontal acceleration would be zero.

v_{x}= u_{x}= u cos θ = 40 m/s

s_{x}= u_{x}t = (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

v_{y}= u_{y}- gt

= u sin θ = gt

= 40 - 20

= 20 m/s

∴

i.e.,

i.e.,

**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/s ^{2})**

or

or

or t^{2}= 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/s ^{2})**

See figure to the answer to question no. 4.

u = 40 m/s, θ = 60°

∴ u_{x}= 40 cos 60° = 20 m/s

Thus, v_{x}= u_{x}= 20 m/s

As φ = 45°

∴ y_{y}= v_{x}= 20 m/s

Thus,

= 20√2 m/s

Before reaching highest point

v_{y}= u_{y}+ (- g) t∴ 20 = 40 sin 60° - 10 t

or

⇒

After attaining highest point

i.e.,

or

**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

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

Change in velocity

= (- 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 H

_{max}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 ****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

y-coordinate will be zero when 4t - 5t^{2}= 0i.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

**Q.11. A particle is projected at an angle 60° 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.**

v

_{x}= u_{x}= 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,

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