Question:

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R. 
Assertion (A): Range of a horizontal projectile is maximum when angle of projection is \(θ = 45°\).
Reason (R): Range is maximum when \(sin\ (2θ)=1\).
In the light of the above statements, choose the correct answer from the options given below

Updated On: Mar 20, 2025
  • (A) and (R) both are true and (R) is correct explanation of (A)

  • (A) and (R) both are true but (R) is not correct explanation of (A)

  • (A) is true and (R) is false

  • Both (A) and (R) are false

Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Approach Solution - 1

We know that:

Range \(R = \frac{\mu^2}{g}\,sin\,2\theta\)

R is maximum for \(2\theta=90^o\)

⇒ \(\theta=45^o\)

So, the correct option is (A): (A) and (R) both are true and (R) is correct explanation of (A).

Was this answer helpful?
1
0
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

Projectile Motion Range Problem 

Step 1: Horizontal Range Formula

For a ground-to-ground projectile, the horizontal range (R) is given by:

\( R = \frac{u^2 \sin 2\theta}{g} \)

where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

Step 2: Maximum Range Condition

For a given initial velocity (\( u \)), the range (R) is maximum when \( \sin 2\theta \) is maximum. The maximum value of \( \sin 2\theta \) is 1.

Step 3: Angle for Maximum Range

Since the maximum value of \( \sin 2\theta \) is 1, we have:

\( \sin 2\theta = 1 \)

\( 2\theta = \frac{\pi}{2} \)

\( \theta = \frac{\pi}{4} \text{ radians} = 45^\circ \)

Step 4: Analyze Assertion and Reason

Assertion A states that the range is maximum when the projection angle is 45°. This is correct, as shown in Step 3.

Reason R states that for maximum range, \( \sin 2\theta \) should be equal to one. This is also correct, as shown in Step 2.

Furthermore, Reason R explains Assertion A because the maximum value of \( \sin 2\theta \) (which is 1) occurs when \( \theta = 45^\circ \), leading to the maximum range.

Conclusion:

Both Assertion A and Reason R are correct, and Reason R is the correct explanation of Assertion A (Option 1).

Was this answer helpful?
0
0

Concepts Used:

Projectile Motion

Projectile

A projectile is an object set in flight by applying an external force. The projectile may be thrown up vertically or at an angle to the horizontal. It may be dropped from a position of rest. The only force acting on the projectile during its motion along the flight path is the gravitational force and it is in motion due to its own inertia

Examples of projectile are: a ball hit by a bat, bullet fired from a gun, shell launched from a launcher, bomb dropped from a plane, etc. It must be noted that a rocket or a missile cannot be considered as a projectile as they are propelled by power.

Trajectory path/ projectile motion

When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the centre of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the motion is called projectile motion or trajectory motion.

In a Projectile Motion, there are two simultaneous independent rectilinear motions:

  1. Along the x-axis: uniform velocity, responsible for the horizontal (forward) motion of the particle.
  2. Along y-axis: uniform acceleration, responsible for the vertical (downwards) motion of the particle.

Acceleration in the horizontal projectile motion and vertical projectile motion of a particle: When a particle is projected in the air with some speed, the only force acting on it during its time in the air is the acceleration due to gravity (g). This acceleration acts vertically downward. There is no acceleration in the horizontal direction, which means that the velocity of the particle in the horizontal direction remains constant.

The types of Projectile Motion Formula are: 

  • Horizontal Distance – x = Vx0t
  • Horizontal Velocity – Vx = Vx0
  • Vertical Distance, y – Vy0t – ½ gt2
  • Vertical Velocity, Vy – Vy0 – gt