Problem 18 - Master Resource Book in JEE Main Physics

Question

Two projectiles \(A\) and \(B\) thrown with speeds in the ratio \(1: \sqrt{2}\) acquired the same heights. If \(A\) is thrown at an angle of \(45^{\circ}\) with the horizontal, the angle of projection of \(B\) will be (a) \(0^{\circ}\) (b) \(60^{\circ}\) (c) \(30^{\circ}\) (d) \(45^{\circ}\) (e) \(15^{\circ}\)

Step-by-Step Solution

Answer

Projectile B is thrown at an angle of 30°, i.e., option (c).

Step 1: Understanding the Problem

We need to find the angle of projection for projectile B given that both projectiles A and B reach the same maximum height. Projectile A is thrown at 45 degrees and with a speed that is in ratio 1 to \(\sqrt{2}\) compared to projectile B's speed.

Step 2: Use the Maximum Height Formula

The maximum height \(H\) of a projectile can be given by the formula: \[ H = \frac{{v^2 \sin^2 \theta}}{{2g}} \] where \(v\) is the initial velocity, \(\theta\) is the projection angle, and \(g\) is the acceleration due to gravity. Since both projectiles reach the same height \(H_A = H_B\).

Step 3: Set Up the Equations for Projectile A and B

For projectile A: \[ H_A = \frac{{v_A^2 \sin^2 45}}{{2g}} \] Since \(\sin 45 = \frac{\sqrt{2}}{2}\), simplify as: \[ H_A = \frac{{v_A^2 \cdot \left( \frac{\sqrt{2}}{2} \right)^2}}{{2g}} = \frac{{v_A^2}}{{4g}} \]For projectile B: \(v_B = \sqrt{2}v_A \), \[ H_B = \frac{{v_B^2 \sin^2 \theta}}{{2g}} = \frac{{2v_A^2 \sin^2 \theta}}{{2g}} = \frac{{v_A^2 \sin^2 \theta}}{{g}} \]

Step 4: Equate Maximum Heights for A and B

Since \(H_A = H_B\), we equate: \[ \frac{{v_A^2}}{{4g}} = \frac{{v_A^2 \sin^2 \theta}}{{g}} \]Canceling \(v_A^2/g\) from both sides gives: \[ \frac{1}{4} = \sin^2 \theta \]

Step 5: Solve for \(\theta\)

Taking the square root of both sides results in: \[ \sin \theta = \frac{1}{2} \]This implies that \( \theta = 30^{\circ} \), since \( \sin 30^{\circ} = \frac{1}{2} \).

Step 6: Conclusion: Choose the Correct Answer

The angle of projection for projectile B is \(30^{\circ}\), which matches with option (c).

Key Concepts

Maximum Height

The maximum height reached by a projectile is an essential part of studying projectile motion. It refers to the highest point a projectile achieves before it starts descending. This point happens when the vertical component of the velocity becomes zero. To understand this, we use the formula for maximum height:\[ H = \frac{{v^2 \sin^2 \theta}}{{2g}} \]Here, \(v\) stands for initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity. The formula highlights how the sine of the angle and the square of the initial velocity directly influence the maximum height. It's important because achieving the same maximum height, as seen in our exercise, indicates a balance between these factors for different projectiles.

Angle of Projection

The angle of projection is the angle at which a projectile is launched, relative to the horizontal. It plays a crucial role in determining the path and behavior of the projectile. In our exercise, projectile A was launched at an angle of \(45^\circ\). This angle is significant because it typically results in maximum range for a given initial velocity.

• A steeper angle means higher maximum height but shorter range.
• A shallower angle results in a longer range but lower maximum height.

The solution required finding the angle for projectile B, which was determined using the relationship between the initial velocities and maximum height. It was calculated to be \(30^\circ\), demonstrating how different angles can result in the same height when initial velocities vary.

Initial Velocity

Initial velocity is the speed at which a projectile is launched. It's a vector quantity, meaning it has both magnitude and direction. In the problem at hand, the initial velocities of the two projectiles were related by a ratio.To analyze motion, it's useful to break initial velocity into horizontal and vertical components:

• Horizontal component: \(v_0 \cos \theta\)
• Vertical component: \(v_0 \sin \theta\)

These components influence how high and how far the projectile will travel. Understanding the ratio of initial velocities helped establish the conditions for both projectiles achieving the same height. For projectile B, which was faster, the increased velocity compensated for the decreased angle to attain the same maximum height as projectile A.

Acceleration due to Gravity

Acceleration due to gravity, denoted as \(g\), is approximately \(9.8 \text{ m/s}^2\) on Earth's surface. It's a constant force acting downwards that affects all objects in projectile motion. Understanding this natural force is key to analyzing vertical motion, as it determines how a projectile will decelerate upwards and accelerate downwards.Here’s how acceleration due to gravity impacts projectile motion:

• It influences the time a projectile spends in the air.
• It affects the vertical velocity, altering the speed and direction over time.

In our exercise, \(g\) was an important factor in calculating maximum height. It worked with the other variables to ensure that the equation accounted for how gravity influences the vertical journey of the projectile, ultimately helping us solve for the angle of projection for projectile B.