Problem 53 - Master Resource Book in JEE Main Physics

Question

The maximum height attained by a projectile when thrown at an angle \(\theta\) with the horizontal is found to be half the horizontal range. Then, \(\theta\) is equal to [KCET 2007] (a) \(\tan ^{-1}(2)\) (b) \(\frac{\pi}{6}\) (c) \(\frac{\pi}{4}\) (d) \(\tan ^{-1}\left(\frac{1}{2}\right)\)

Step-by-Step Solution

Answer

The angle \( \theta \) is \( \frac{\pi}{4} \).

Step 1: Analyze Given Information

We know that the maximum height, H, attained by a projectile is equal to half of the horizontal range, R. This means we set up the equation \( H = \frac{1}{2}R \).

Step 2: Recall the Formula for Maximum Height

The maximum height \( H \) of a projectile launched with initial velocity \( v \) at an angle \( \theta \) is given by: \[ H = \frac{v^2 \sin^2 \theta}{2g} \]where \( g \) is the acceleration due to gravity.

Step 3: Recall the Formula for Horizontal Range

The horizontal range \( R \) is given by: \[ R = \frac{v^2 \sin 2\theta}{g} \]This relates the initial velocity \( v \), angle of projection \( \theta \), and acceleration due to gravity \( g \).

Step 4: Set up the Equation Based on Condition

From Step 1, we have that \( H = \frac{1}{2}R \). Substitute \( H \) and \( R \) from step 2 and 3:\[ \frac{v^2 \sin^2 \theta}{2g} = \frac{1}{2} \left( \frac{v^2 \sin 2\theta}{g} \right) \]

Step 5: Simplify and Solve for \( \theta \)

Cancel \( v^2 \) and \( g \) from both sides, then simplify:\[ \sin^2 \theta = \frac{1}{2} \sin 2\theta \]Since \( \sin 2\theta = 2 \sin\theta \cos\theta \), substitute:\[ \sin^2 \theta = \sin\theta \cos\theta \]Divide both sides by \( \sin\theta \) (assuming \( \sin\theta eq 0 \)):\[ \sin\theta = \cos\theta \]which implies that \( \tan\theta = 1 \).

Step 6: Determine the Angle \( \theta \)

From \( \tan\theta = 1 \), we find that \( \theta = \tan^{-1}(1) = \frac{\pi}{4} \) radians.

Key Concepts

Maximum Height

In projectile motion, the maximum height refers to the highest vertical position reached by a projectile in its flight. To understand this better, let's consider the motion of an object thrown into the air at an angle. The maximum height is reached when the object's vertical velocity becomes zero, right before it starts to descend back to the ground. You can think of it as the peak or the top of the arc.

The formula to calculate the maximum height, given an initial velocity (v) and the angle of projection (\theta), is:

• \[ H = \frac{v^2 \sin^2 \theta}{2g} \]

Here:

• \( H \) is the maximum height.
• \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth).

This formula explains how the initial speed and the angle at which you throw the projectile affect its ability to reach a high point. Understanding this concept is crucial in many real-world applications, such as sports, engineering, and even video game design, where predicting the trajectory of a projectile can be essential.

Horizontal Range

The horizontal range of a projectile is the total distance traveled by the projectile along the horizontal axis before touching the ground. It is a key concept when analyzing the motion of any thrown or launched object. In simple terms, this is how far your projectile reaches if you throw it in a straight line along the ground

To calculate horizontal range, we use the following formula:

• \[ R = \frac{v^2 \sin 2\theta}{g} \]

In this equation:

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

This formula shows that both the speed and the angle of projection are crucial in determining how far a projectile can go. For any given initial speed, there's a specific angle that maximizes the range, commonly discussed in physics problems.

Angle of Projection

The angle of projection in projectile motion is the angle at which an object is launched with respect to the horizontal axis. This angle greatly influences both the maximum height attained by the projectile and the horizontal range it covers during its motion.

When analyzing these motions, the angle of projection is important because:

• It affects how gravity acts on the projectile, determining its path or trajectory.
• This angle influences how far and how high the projectile goes.

It is interesting to note that for a fixed initial speed, the horizontal range is maximized when the angle of projection is 45 degrees, or \( \frac{\pi}{4} \) radians. This is explained by the fact that when the angle is at this optimal value, the sin and cos terms in the range equation are balanced to give the maximum possible product.

Understanding the angle of projection is integral not only in academic scenarios but also in practical applications such as sports, where athletes need precise angles to achieve desired distances.