Problem 120 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Two stones are projected from the top of a cliff h metres high, with the same speed \(\mathrm{u}\), so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected horizontally and the other is projected at an angle \(\theta\) to the horizontal then \(\tan \theta\) equals \([2003]\) (a) \(u \sqrt{\frac{2}{g h}}\) (b) \(\sqrt{\frac{2 u}{g h}}\) (c) \(2 g \sqrt{\frac{u}{h}}\) (d) \(2 h \sqrt{\frac{u}{g}}\)

Step-by-Step Solution

Answer

The correct answer is (a) \( u \sqrt{\frac{2}{g h}} \).

Step 1: Understand the Problem

We need to find the expression for \( \tan \theta \) that allows two stones with the same speed \( u \) to hit the same spot on the ground, where one stone is projected horizontally and the other at an angle \( \theta \) from a height \( h \).

Step 2: Analyze Projectile Motion

For the stone projected horizontally, the time \( t_1 \) to reach the ground is determined by the vertical motion, using the formula \( h = \frac{1}{2} g t_1^2 \), solving for \( t_1 \) gives \( t_1 = \sqrt{\frac{2h}{g}} \). The horizontal range is \( R = u t_1 = u \sqrt{\frac{2h}{g}} \).

Step 3: Consider the Stone Projected at Angle \( \theta \)

For the stone projected at an angle \( \theta \), the horizontal component of the velocity is \( u \cos \theta \) and the vertical component is \( u \sin \theta \). The time \( t_2 \) to hit the ground is calculated using the vertical motion: \( h = u \sin \theta \cdot t_2 - \frac{1}{2} g t_2^2 \).

Step 4: Equate the Horizontal Ranges

The horizontal range for the stone projected at an angle is \( R = u \cos \theta \cdot t_2 \). Since both stones hit the same spot, set this equal to \( u \sqrt{\frac{2h}{g}} \).

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

Set \( u \cos \theta \cdot t_2 = u \sqrt{\frac{2h}{g}} \). Simplifying gives \( \cos \theta \cdot (\frac{u \sin \theta + \sqrt{u^2 \sin^2 \theta + 2gh}}{g}) = \sqrt{\frac{2h}{g}} \). Simplifying further can lead to the expression for \( \tan \theta \).

Step 6: Final Calculation and Match with Options

After simplifying, find that \( \tan \theta = u \sqrt{\frac{2}{g h}} \), which matches option (a).

Key Concepts

Horizontal Projection

When an object is launched purely in a horizontal direction from a height, it follows a trajectory due to gravity acting downwards. This type of motion is called horizontal projection. In this case, the initial vertical velocity is zero since the object is only given a horizontal speed.
To determine how long it takes to hit the ground, we consider the vertical motion under gravity. Using the formula for vertical distance:

• \[ h = \frac{1}{2} g t_1^2 \]
• Solving for time \( t_1 \) we get: \( t_1 = \sqrt{\frac{2h}{g}} \)

The horizontal range (distance traveled horizontally) is then given by the product of time and horizontal speed \( u \), such that:

• \( R = u \sqrt{\frac{2h}{g}} \)

Angle of Projection

The angle at which a projectile is launched affects both the range and the height of the trajectory. In cases where the projectile is launched at an angle \( \theta \), it's important to break down the initial velocity \( u \) into horizontal and vertical components:

• Horizontal Component: \( u \cos \theta \)
• Vertical Component: \( u \sin \theta \)

These components are crucial to determine the time of flight and the horizontal range. The angle of projection is a key factor in ensuring both stones hit the same spot. Finding \( \tan \theta \), which is the ratio of vertical to horizontal components, helps compare these motion aspects when another object is projected horizontally.
This leads to identifying the correct trajectory matching both stones, enabling accurate blending of angles and initial speeds.

Range of Projectile

The range of a projectile is the horizontal distance it travels before hitting the ground. In projectile motion, this depends on both the initial speed and the time the projectile is in the air.
For a stone projected from a height with horizontal velocity, the range is given by:

• \( R = u \sqrt{\frac{2h}{g}} \)

For a projectile launched at an angle \( \theta \), the range involves both horizontal component \( u \cos \theta \) and time \( t_2 \). To find this range, equate it to the horizontal motion of the horizontally launched stone, ultimately balancing equal impact points. This systematic approach helps demonstrate how various factors adjust for consistent results.

Time of Flight

Time of flight is the total time a projectile is in the air before it lands. For the horizontally launched stone, this is straightforwardly calculated from vertical motion:

• \( t_1 = \sqrt{\frac{2h}{g}} \)

For a stone launched at an angle, the time of flight \( t_2 \) is determined by both initial velocity components and gravity:

• Solve the vertical motion equation: \( h = u \sin \theta \cdot t_2 - \frac{1}{2} g t_2^2 \)
• Time \( t_2 \) is solved from this quadratic form

The crucial aspect here is ensuring that \( t_2 \) helps equate the horizontal ranges, simplifying into expressions combining initial speeds, the angle \( \theta \), and gravitational force. Time of flight ultimately dictates how long both motions last and ensures aligned results.