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Question

If the horizontal range of a projectile is maximum then the angle of the projectile must be ______ with horizontal.

The correct answer is 45° 

Projectile Motion Fundamentals

Projectile motion describes the path an object takes when thrown or projected into the air, subject only to the force of gravity. Understanding key parameters like initial velocity, angle of projection, and acceleration due to gravity is crucial for analyzing this type of motion.

Horizontal Range in Projectile Motion

The horizontal range of a projectile refers to the total horizontal distance covered by the projectile from its point of projection to the point where it lands on the same horizontal level. For a projectile launched with an initial velocity \(u\) at an angle \(\theta\) with the horizontal, the formula for its horizontal range \(R\) is given by:

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

Where:

  • \(R\) is the horizontal range.
  • \(u\) is the initial velocity of the projectile.
  • \(\theta\) is the angle of projection with the horizontal.
  • \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\) on Earth).

Maximizing Projectile Range

To achieve the maximum horizontal range, we need to consider the formula for \(R\). Since \(u\) (initial velocity) and \(g\) (acceleration due to gravity) are generally constant for a given scenario, the horizontal range \(R\) depends directly on the value of \(\sin(2\theta)\). For \(R\) to be maximum, the term \(\sin(2\theta)\) must be at its maximum possible value.

The maximum value that the sine function, \(\sin(x)\), can achieve is 1. This occurs when the angle \(x\) is \(90^\circ\) (or \(\pi/2\) radians).

Therefore, for the horizontal range \(R\) to be maximum, we must have:

\(\sin(2\theta) = 1\)

This implies that the angle \(2\theta\) must be \(90^\circ\):

\(2\theta = 90^\circ\)

Now, to find the angle of projection \(\theta\), we simply divide by 2:

\(\theta = \frac{90^\circ}{2}\)

\(\theta = 45^\circ\)

Optimal Angle for Maximum Horizontal Range

Thus, for the horizontal range of a projectile to be its maximum, the angle of projection must be 45 degrees with the horizontal. This is a fundamental concept in projectile motion and is often tested in physics examinations.

When a projectile is launched at an angle of 45 degrees, it balances the time it spends in the air (which increases with launch angle) and its initial horizontal velocity component (which decreases with launch angle), leading to the greatest possible horizontal distance travelled.

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Important Questions from Projectiles

  1. The range of a projectile is maximum, when the angle of projection is -

  2. 'If 'θ' is angle of projection and 'u' is velocity of projection for a projectile, then its horizontal range is given by-

  3. A particle is projected with velocity u at an inclination θ with the horizontal. Then Maximum height (H) attained is

  4. The angle of projection of a projectile is _________ when the range will be maximum for a given velocity.

  5. Which of the following is NOT a projectile motion?

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