Problem 51 - Master Resource Book in JEE Main Physics

Question

A particle is projected with certain velocity at two different angles of projections with respect to horizontal plane so as to have same range \(R\) on a horizontal plane. If \(t_{1}\) and \(t_{2}\) are the time taken for the two paths, the which one of the following relations is correct? [UP SEE 2008] (a) \(t_{1} t_{2}=\frac{2 R}{g}\) (b) \(t_{1} t_{2}=\frac{R}{g}\) (c) \(t_{1} t_{2}=\frac{g}{2 g}\) (d) \(t_{1} t_{2}=\frac{4 R}{g}\)

Step-by-Step Solution

Answer

The correct relation is (a) \( t_1 t_2 = \frac{2R}{g} \).

Step 1: Understanding the Problem

We need to determine the relationship between the times of flight, \( t_1 \) and \( t_2 \), for two different angles of projection that result in the same range \( R \). We are given multiple choice options with expressions for \( t_1 t_2 \).

Step 2: Formula for Range in Projectile Motion

The range \( R \) of a projectile is given by the formula \( R = \frac{v^2 \sin(2\theta)}{g} \), where \( v \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity. When two angles give the same range, their angles of projection are complementary, i.e., \( \theta_1 + \theta_2 = 90^\circ \).

Step 3: Understanding Complementary Angles

If two angles \( \theta_1 \) and \( \theta_2 \) are complementary, then the equations for range simplify to account for symmetric properties in trigonometric functions, such that both add up to \( 90^\circ \). Hence, we use \( \sin(2\theta_1) = \cos(2\theta_2) \).

Step 4: Time of Flight Formula

The time of flight \( T \) for a projectile is given by \( T = \frac{2v \sin(\theta)}{g} \). For the two angles of projection \( \theta_1 \) and \( \theta_2 \), the times are \( t_1 = \frac{2v \sin(\theta_1)}{g} \) and \( t_2 = \frac{2v \sin(\theta_2)}{g} \).

Step 5: Relation between \( t_1 \) and \( t_2 \)

Since \( \theta_1 + \theta_2 = 90^\circ \), \( \sin(\theta_1) = \cos(\theta_2) \) and \( \sin(\theta_2) = \cos(\theta_1) \). This means \( \sin(\theta_1)\sin(\theta_2) = \sin(\theta_1) \cos(\theta_1) = \frac{1}{2} \sin(2\theta_1) \).

Step 6: Calculating \( t_1 t_2 \)

\( t_1 t_2 = \left(\frac{2v \sin(\theta_1)}{g}\right) \left(\frac{2v \sin(\theta_2)}{g}\right) = \frac{4v^2 \sin(\theta_1) \sin(\theta_2)}{g^2} \). Since \( \sin(\theta_1)\sin(\theta_2) = \frac{1}{2} \sin(2\theta_1) \), we have \( t_1 t_2 = \frac{4v^2 (\frac{1}{2} \sin(2\theta_1))}{g^2} = \frac{2R}{g} \).

Step 7: Conclusion

From the derived expression, \( t_1 t_2 = \frac{2R}{g} \), which corresponds to option (a).

Key Concepts

Time of Flight

In projectile motion, the time of flight refers to the total time a projectile takes to hit the ground after it is launched. This can be calculated using the formula: \[ T = \frac{2v \sin(\theta)}{g} \]Where:

• \( T \) is the time of flight.
• \( v \) is the initial velocity of the projectile.
• \( \theta \) is the angle of projection.
• \( g \) is the acceleration due to gravity, usually approximated as \( 9.8 \text{ m/s}^2 \).

For different angles with the same initial speed, the calculated times will vary. Interestingly, if you change the angle and keep the speed constant, the time will increase or decrease depending on how high the projectile needs to travel. If two angles are used to achieve the same horizontal distance (range), their respective times of flight, labeled as \( t_1 \) and \( t_2 \), can be related using specific complementary angle properties in trigonometry.

Range of Projectile

The range of a projectile is essentially how far the projectile travels horizontally through the air. To find this range, you can rely on a key formula in projectile motion:\[ R = \frac{v^2 \sin(2\theta)}{g} \]Where:

• \( R \) is the range.
• \( v \) is the initial velocity.
• \( \theta \) is the angle of projection.
• \( g \) is the acceleration due to gravity.

A fascinating property occurs when two projectiles are launched at complementary angles, meaning their angles add up to \( 90^\circ \). Both projectiles will cover the same range \( R \), even though their flight paths and times of flight differ. Understanding this can be essential in solving complex problems involving varying angles of projection, and it allows for strategic planning in scenarios requiring accurate projectile placements.

Complementary Angles

Complementary angles in the context of projectile motion are pairs of angles that add up to \( 90^\circ \). For projectiles, using these pairs of angles is a clever way to achieve a specific range while varying the projectile's flight path and time of flight. When using complementary angles:

• The sine and cosine functions have symmetrical properties: \( \sin(\theta) = \cos(90^\circ - \theta) \) and vice versa.
• Both projectiles launched at these angles will have the same range due to the symmetry in trigonometric values.
• They possess different times of flight, \( t_1 \) and \( t_2 \), leading to different durations via different paths to achieve the same end point.

Harnessing the concept of complementary angles is critical when you need to manipulate the path of a projectile to achieve precise distances, impacts, or travel times. This concept simplifies the involved calculations by utilizing inherent symmetries in mathematical trigonometry.