Problem 38 - Master Resource Book in JEE Main Physics

Question

The horizontal range of an oblique projectile is equal to the distance through which a projectile has to fall freely from rest to acquire a velocity equal to the velocity of projection in magnitude. The angle of projection is (a) \(15^{\circ}\) (b) \(60^{*}\) (c) \(45^{\circ}\) (d) \(3 \overline{0^{*}}\)

Step-by-Step Solution

Answer

The angle of projection is 15°.

Step 1: Understand the Problem

We need to find the angle of projection such that the horizontal range of an oblique projectile is equal to the vertical distance it would fall freely under gravity to reach the same speed as the initial speed of projection.

Step 2: Identify Key Equations and Variables

For a projectile, the horizontal range is given by \(R = \frac{v^2 \sin(2\theta)}{g}\), where \(v\) is the initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity. The vertical distance \(h\) can be found using \(v^2 = 2gh\), giving \(h = \frac{v^2}{2g}\).

Step 3: Set the Horizontal Range Equal to the Vertical Distance

Since the horizontal range \(R\) equals the vertical fall distance \(h\), we equate the two: \[\frac{v^2 \sin(2\theta)}{g} = \frac{v^2}{2g}\].

Step 4: Simplify the Equation

Cancel \(v^2\) and \(g\) from both sides of the equation to get \(\sin(2\theta) = \frac{1}{2}\).

Step 5: Solve for the Angle \(\theta\)

The equation \(\sin(2\theta) = \frac{1}{2}\) implies \(2\theta = 30^{\circ}\) or \(2\theta = 150^{\circ}\). Solving for \(\theta\), we find \(\theta = 15^{\circ}\) or \(\theta = 75^{\circ}\). However, only \(\theta = 15^{\circ}\) matches the options provided.

Step 6: Verify the Answer

Since both angles \(15^{\circ}\) and \(75^{\circ}\) satisfy the condition \(\sin(2\theta) = \frac{1}{2}\), \(\theta = 15^{\circ}\) is the correct choice under the given options.

Key Concepts

Horizontal Range

The horizontal range of a projectile is the distance it travels horizontally during its flight. This depends on the initial velocity, the projection angle, and the effect of gravity.

• The formula for horizontal range is given by \[ R = \frac{v^2 \sin(2\theta)}{g} \] where \(v\) is the initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity.
• The function \(\sin(2\theta)\) is crucial here, indicating that the range will vary as the angle changes.
• A specific angle, \(45^{\circ}\), maximizes the range when air resistance is negligible.

Understanding horizontal range helps predict where a projectile will land, which is useful in activities ranging from sports to engineering.

Angle of Projection

The angle of projection is the angle at which a projectile is launched relative to the horizontal axis. This angle plays a significant role in determining the trajectory and the range of the projectile.

• The angle affects both the vertical and horizontal components of the initial velocity.
• The equation \(\sin(2\theta)\) shows the importance of the angle in maximizing the horizontal range.
• In many cases, particularly when no air resistance is considered, a \(45^{\circ}\) angle yields maximum range.

Understanding the optimal angle of projection can be crucial in applications like artillery, sports, and even launching spacecraft. Practicing different angles affects both the speed and distance the projectile travels.

Free Fall Distance

Free fall distance refers to the vertical distance a projectile or object falls under the influence of gravity alone, starting from rest. This concept is important in comparing a projectile's horizontal range to its vertical fall distance.

• The formula to calculate free fall distance is \[ h = \frac{v^2}{2g} \] where \(v\) is the velocity it achieves, and \(g\) is the acceleration due to gravity.
• This vertical distance \(h\) is where the projectile will naturally reach the same speed as the initial speed of the projectile if it were in free fall.
• It helps us relate the projectile motion back to a vertical drop, offering a clear understanding of both motion types.

Comparing free fall distance with horizontal range allows for unique insights into motion dynamics and helps in determining projection angles for desired target reach.

Acceleration Due to Gravity

Acceleration due to gravity, denoted as \(g\), is a crucial constant in physics that describes how strongly objects accelerate toward Earth when they fall freely. Its value is approximately \(9.8 \, \text{m/s}^2\) near the Earth's surface.

• This constant appears in many projectile motion equations, such as the ones for horizontal range and free fall distance.
• Gravity impacts the projectile's vertical motion, pulling it downwards and influencing its trajectory curves.
• In uniform gravitational fields, all objects fall with the same acceleration if air resistance is negligible.

Understanding acceleration due to gravity is essential for solving problems involving free fall and projectile motions. It is consistent across similar scenarios but may vary slightly with altitude or location due to Earth's geometry.