Problem 23 - Master Resource Book in JEE Main Physics

Question

Two particles are projected in air with speed \(v_{0}\) at angles \(\theta_{1}\) and \(\theta_{2}\) (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices (a) angle of projection : \(\theta_{1}>\theta_{2}\) (b) time of flight : \(T_{1}>T_{2}\) (c) horizontal range : \(R_{1}>R_{2}\) (d) total energy : \(U_{1}>U_{2}\)

Step-by-Step Solution

Answer

(a) and (b) are correct.

Step 1: Analyzing Height of Projection

The maximum height reached by a projectile is given by the formula \( H = \frac{v_{0}^{2} \sin^{2} \theta}{2g} \). Given that the height reached by the first particle is greater than that of the second, we have: \[ \frac{v_{0}^{2} \sin^{2} \theta_{1}}{2g} > \frac{v_{0}^{2} \sin^{2} \theta_{2}}{2g} \] This simplifies to: \[ \sin^{2} \theta_{1} > \sin^{2} \theta_{2} \] Therefore, \( \sin \theta_{1} > \sin \theta_{2} \), meaning \( \theta_{1} > \theta_{2} \). Thus, option (a) \(\theta_{1} > \theta_{2} \) is correct.

Step 2: Comparing Time of Flight

Time of flight for a projectile is given by \( T = \frac{2v_{0} \sin \theta}{g} \). The time of flight for the first particle is \( T_{1} = \frac{2v_{0} \sin \theta_{1}}{g} \) and for the second particle \( T_{2} = \frac{2v_{0} \sin \theta_{2}}{g} \). Based on \( \sin \theta_{1} > \sin \theta_{2} \), we have \( T_{1} > T_{2} \). Thus, option (b) \( T_{1} > T_{2} \) is correct.

Step 3: Evaluating Horizontal Range

The horizontal range of a projectile is given by \( R = \frac{v_{0}^{2} \sin 2\theta}{g} \). Here, \( R_{1} = \frac{v_{0}^{2} \sin 2\theta_{1}}{g} \) and \( R_{2} = \frac{v_{0}^{2} \sin 2\theta_{2}}{g} \). Since \( \sin 2\theta_{1} < \sin 2\theta_{2} \) for acute angles with \( \theta_{1} > \theta_{2} \), hence \( R_{1} < R_{2} \). Therefore, option (c) \( R_{1} > R_{2} \) is incorrect.

Step 4: Assessing Total Energy

The total mechanical energy of the projectile is the same for both particles because it depends only on the initial speed and the mass of the particles, provided no external forces act on them besides gravity. Thus, option (d) \( U_{1} > U_{2} \) is incorrect.

Key Concepts

Maximum Height

In projectile motion, the maximum height refers to the tallest vertical point a projectile reaches during its trajectory. This height depends on the vertical component of the initial velocity and the angle of projection. The formula used for calculating maximum height is \[ H = \frac{v_{0}^{2} \sin^{2} \theta}{2g} \]where:

• \( v_{0} \) is the initial velocity,
• \( \theta \) is the angle of projection,
• \( g \) is the acceleration due to gravity.

When comparing two projectiles, if one reaches a greater height, it means that the sine of its angle of projection is larger. Given that \( \sin \theta_{1} > \sin \theta_{2} \), the angle \( \theta_{1} \) must be greater than \( \theta_{2} \) for acute angles.

Time of Flight

Time of flight refers to the total time a projectile is in motion from the point of launch until it reaches the ground again. It is determined by the vertical component of the initial velocity and is calculated using the formula: \[ T = \frac{2v_{0} \sin \theta}{g} \]where:

• The terms in the formula remain consistent with those in the maximum height calculation.

If \( \theta_{1} > \theta_{2} \), then \( \sin \theta_{1} > \sin \theta_{2} \). This leads to a longer time of flight for the projectile launched at \( \theta_{1} \) when compared with that at \( \theta_{2} \). Therefore, the time of flight \( T_{1} \) is greater than \( T_{2} \).

Horizontal Range

The horizontal range of a projectile is the total horizontal distance traveled during its motion. It depends on the angle of projection and both horizontal and vertical components of the initial velocity. The formula for calculating the horizontal range is:\[ R = \frac{v_{0}^{2} \sin 2\theta}{g} \]Here:

• \( \sin 2\theta \) accounts for the interaction between the vertical and horizontal components.

For two projectiles with angles \( \theta_{1} > \theta_{2} \), the value of \( \sin 2\theta \) does not increase linearly. In fact, for acute angles and given \( \theta_{1} > \theta_{2} \), it means \( R_{1} < R_{2} \). Hence, the projectile with a smaller angle has a larger horizontal range.

Mechanical Energy

Mechanical energy in projectile motion is the sum of the potential and kinetic energies. It is dependent on the initial speed and mass of the projectile and remains constant if only gravity acts on it and no other external forces are involved. The mechanical energy can be represented as:\[ E = \frac{1}{2} mv_{0}^{2} \]where:

• \( m \) is the mass of the projectile,
• \( v_{0} \) remains the initial velocity.

Since both projections have the same initial speed and assuming they have the same mass, their mechanical energies remain equal throughout their flight, regardless of their individual trajectories or angles of projection.