Problem 40 - Master Resource Book in JEE Main Physics

Question

Two stones are projected so as to reach the same distance from the point of projection on a horizontal surface. The maximum height reached by one exceeds the other by an amount equal to half the sum of the height attained by them. Then, angle of projection of the stone which attains smaller height is [a] \(45^{\circ}\) (b) \(60^{\circ}\) (c) \(3 \overline{0^{*}}\) (d) \(\tan ^{-1}(3 / 4)\)

Step-by-Step Solution

Answer

The angle of projection of the stone with smaller height is \( \tan^{-1}(3/4)\), option (d).

Step 1: Understanding the Problem

We have two stones projected at different angles but achieving the same horizontal range. Let's denote the heights reached by them as \(h_1\) and \(h_2\), with \(h_1 > h_2\). We know \(h_1 = h_2 + \frac{1}{2}(h_1 + h_2)\). We need to determine the angle of projection for the stone reaching the smaller height \(h_2\).

Step 2: Relate Maximum Height to Projection Angle

The maximum height for projectile motion is given by \(h = \frac{u^2 \sin^2 \theta}{2g}\), where \(u\) is the initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity. Let \(u_1\) and \(\theta_1\) be for the stone that reaches \(h_1\), and \(u_2\) and \(\theta_2\) for the stone that reaches \(h_2\).

Step 3: Equal Range Condition

The range \(R\) for both stones is given by \(R = \frac{u^2 \sin(2\theta)}{g}\). Since both stones reach the same range, \( \frac{u_1^2 \sin(2\theta_1)}{g} = \frac{u_2^2 \sin(2\theta_2)}{g} \), implying \( u_1 \sin(2\theta_1) = u_2 \sin(2\theta_2) \).

Step 4: Use the Given Relationship Between Heights

From the problem statement, we have the equation \(h_1 = h_2 + \frac{1}{2}(h_1 + h_2)\). This simplifies to \(2h_1 = 3h_2\), giving \(h_1 = \frac{3}{2}h_2\). Substitute the values of \(h_1\) and \(h_2\) using their formulas.\[\frac{u_1^2 \sin^2 \theta_1}{2g} = \frac{3}{2} \frac{u_2^2 \sin^2 \theta_2}{2g}\].

Step 5: Determine the Angle for the Smaller Height

Combine the height relationship and range equation: \(\frac{u_1^2 \sin^2 \theta_1}{u_2^2 \sin^2 \theta_2} = \frac{3}{2}\) and \(u_1 \sin(2\theta_1) = u_2 \sin(2\theta_2)\). Solving these, we find \(\tan^2(\theta_2) = \frac{3}{4}\), which leads to \( \theta_2 = \tan^{-1}\left(\frac{3}{4}\right)\).

Step 6: Conclusion: Select the Appropriate Answer

Based on our calculations, the angle of projection for the stone with the smaller height (\(h_2\)) is \( \tan^{-1}\left(\frac{3}{4}\right)\), which matches option (d).

Key Concepts

Angle of Projection

The angle of projection in projectile motion determines the initial direction with respect to the horizontal. Imagine you are tossing a ball. If you throw it at a steep angle, the ball goes higher but doesn't travel very far horizontally. Conversely, a lower angle means it doesn’t reach as high but travels longer horizontally.
The angle significantly influences both the maximum height and the horizontal range.When dealing with projectiles achieving the same range but different heights, like in the original problem, the angles of projection differ. If you consider the equations of motion, the height attained by a projectile can be linked to the angle by the equation:\[h = \frac{u^2 \sin^2 \theta}{2g}\] Here, \(u\) is the velocity, \(\theta\) is the angle of projection, and \(g\) is gravity. The "sin" function of the angle plays a crucial role; it increases until \(45^{\circ}\) and then begins to decrease, which shows why angles equal distance on ranges beyond \(45^{\circ}\). It’s fascinating that two different angles can lead to the same range, as shown in the solution above, provided their sum is \(90^{\circ}\).
When solving problems like the one above, converting derived relationships into trigonometric functions - as shown with \( \tan^{-1} \left( \frac{3}{4} \right) \) - helps in arriving at the angles easily and accurately.

Maximum Height

The maximum height of a projectile is the apex of its trajectory - the highest vertical position it achieves. This concept is crucial in deciding how far and how high the projectile will travel.
In physics, the maximum height can be calculated using:\[ h = \frac{u^2 \sin^2 \theta}{2g} \]This equation reveals that

• The initial speed \(u\)
• The angle of projection \(\theta\)
• Gravity \(g\)

are the three critical factors affecting the height. A larger angle usually results in a higher maximum height but may reduce the horizontal distance.
In the exercise shared above, one stone achieves a greater height than its counterpart due to a difference in these factors, corresponding especially to the angle of projection. The relationship between the heights of the stones is expressed as \(h_1 = \frac{3}{2}h_2\). This means that the first stone's peak is 1.5 times higher than the second one, revealing the vital impact of height on projectiles depending on launch conditions, confirming the importance of precisely calculating for these differing conditions using proper trigonometric identities and formulas.

Horizontal Range

The horizontal range is how far a projectile moves horizontally from its launch point to landing. In simpler terms: it's the distance covered on the ground.
The range equation is:\[ R = \frac{u^2 \sin(2\theta)}{g} \]This focuses on initial speed \(u\), gravity \(g\), and the nuanced function \(\sin(2\theta)\). Angle is impactful here, as \(\sin(2\theta)\) accounts for the symmetrical relationship in projection angles.For instance, a \(30^{\circ}\) and a \(60^{\circ}\) angle yields the same range as they both feature sums leading to \(90^{\circ}\).
A key point in the exercise is that both stones achieve the same horizontal range despite different maximum heights. This demonstrates one of projectile motion's intriguing principles: the same range can stem from distinct angles if their sine products equal. It showcases the significance of angle in tailoring launches to desired outcomes, illustrating the importance of balancing vertical and horizontal conditions when projecting objects across distances. By cracking these into mathematical forms and translating problems into meaningful functions, students grasp how motion naturally balances height and range in projectile motion.