Problem 1 - Master Resource Book in JEE Main Physics

Question

The horizontal range of a projectile fired at an angle of \(15^{\circ}\) is \(50 \mathrm{~m}\). If it is fired with the same speed at an angle of \(45^{\circ}\), its range will be (a) \(60 \mathrm{~m}\) (b) \(\underline{71} \bar{m}\) (c) \(100 \mathrm{~m}\) (d) \(141 \mathrm{~m}\)

Step-by-Step Solution

Answer

The range at 45° is 100 m, option (c).

Step 1: Understanding the Formula for Range

The horizontal range of a projectile is given by the formula: \[ R = \frac{v^2 \sin(2\theta)}{g} \] where \( R \) is the range, \( v \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity (approximately \( 9.81\, \text{m/s}^2 \)).

Step 2: Relating Range with Initial Velocity

For the angle \( 15^{\circ} \), the formula becomes \[ R_{15} = \frac{v^2 \sin(30^{\circ})}{g} \] Given that \( R_{15} = 50 \text{ m} \), we have: \[ 50 = \frac{v^2 \times \frac{1}{2}}{g} \Rightarrow v^2 = \frac{100g}{1} \]

Step 3: Calculating the Range at 45 Degrees

Using the same initial speed at an angle of \( 45^{\circ} \), we update the angle in the range formula: \[ R_{45} = \frac{v^2 \sin(90^{\circ})}{g} \] Since \( \sin(90^{\circ}) = 1 \), it simplifies to \[ R_{45} = \frac{v^2}{g} \] From Step 2, we know \( v^2 = 100g \), therefore: \[ R_{45} = \frac{100g}{g} = 100 \text{ m} \]

Step 4: Selecting the Correct Answer

Now we compare the calculated range \( R_{45} = 100 \text{ m} \) with the options given. The correct answer is (c) \( 100 \) meters.

Key Concepts

Horizontal Range

In the world of physics, especially when dealing with projectile motion, understanding the horizontal range is crucial. The horizontal range refers to the total distance that a projectile will travel along a horizontal plane before hitting the ground.
The horizontal range can be calculated using the formula: - \[ R = \frac{v^2 \sin(2\theta)}{g} \] Where:

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

This formula helps us determine how far a projectile will reach horizontally when launched at a specific speed and angle. Adjusting either the angle or the initial speed can greatly affect the projectile's range, as seen in the exercise where changing the angle from \( 15^{\circ} \) to \( 45^{\circ} \) significantly increased the range.

Angle of Projection

The angle at which a projectile is launched, known as the angle of projection, significantly affects its trajectory and range. In simple terms, this is the angle between the initial velocity vector and the horizontal plane.
Understanding the role of the angle of projection is crucial:

• An angle of \(45^{\circ}\) is often touted as the optimal angle for maximizing horizontal range under ideal conditions.
• Smaller angles, like \(15^{\circ}\), result in a lower trajectory and, generally, a shorter range for the same initial velocity.
• Larger angles, such as \(75^{\circ}\), may send the projectile higher into the air, but typically at the expense of horizontal distance.

Thus, by adjusting the angle of projection, one can control whether the projectile travels a longer horizontal distance or reaches a greater height.

Initial Velocity

Initial velocity is the speed at which a projectile is launched. It is a vector quantity, meaning it has both magnitude and direction.
The initial velocity (\( v \)) directly influences both the height and range of a projectile:

• Greater initial velocity can increase the range of the projectile, given a constant angle of projection.
• It plays a crucial role in determining the overall trajectory, combining with the angle of projection to define the path the projectile will follow.
• In practical applications, initial velocity will include speed in both horizontal and vertical components.

This concept is vital for predicting the success of sports plays, military artillery, and even space missions, where precision in speed and direction can lead to different outcomes.

Acceleration due to Gravity

Acceleration due to gravity is a constant force that acts on any object moving through the Earth’s atmosphere, pulling it down towards the center of the Earth. It is always directed downwards and has a standard value of approximately \(9.81 \, \text{m/s}^2\).
In the context of projectile motion:

• Gravity affects the vertical component of a projectile's motion by causing it to decelerate as it rises and accelerate as it falls.
• The acceleration due to gravity does not affect the horizontal motion directly as there is no acceleration in the horizontal plane once the object is in motion.
• In the range formula \( R = \frac{v^2 \sin(2\theta)}{g} \), gravity acts as the divisor, inversely affecting the horizontal range; increasing gravity would decrease the range, and vice-versa.

Understanding gravity is essential for accurately predicting how far a projectile can travel and helps in engineering solutions to take advantage of or counteract its effects.