# Tried the Last Question from India's Hardest Exam

TLDRThis video script tackles a complex physics problem from the JEE Advanced paper involving a projectile launched at an angle with velocity 'V' under gravity 'G'. The challenge arises when gravity changes at the projectile's peak, altering its range. The presenter methodically calculates the projectile's range in two distinct phases, applying principles of motion and energy conservation, ultimately deriving a formula to find the new range 'D primed' as a multiple 'n' of the original 'D'. The engaging explanation unravels the problem, revealing 'n' to be 0.95, offering viewers a deeper understanding of projectile dynamics under varying gravitational conditions.

###### Takeaways

- π The problem involves a projectile launched at an angle Theta and speed V with gravitational acceleration G, which changes at the projectile's highest point.
- π The original range of the projectile under normal conditions is D.
- π The gravitational acceleration changes to G/0.81 at the highest point, affecting the projectile's range.
- π’ The new range, D', is a multiple of the original range D, and the goal is to find this multiple, denoted as n.
- π The time of flight is crucial for calculating the first part of the range, D1, and is derived from the vertical velocity equation.
- π‘ The horizontal range D1 is calculated using the constant horizontal velocity, V cos Theta, and the previously found time of flight.
- π The total original range D is twice D1 due to the symmetry of projectile motion under constant gravity.
- π The maximum height of the projectile is calculated using the conservation of energy principle and the initial vertical velocity.
- π½ The projectile's descent under the new gravitational acceleration is analyzed using a modified version of the displacement equation.
- π° The time of flight under the new acceleration is calculated, considering the change in gravitational force.
- π The second half of the projectile's range, D2', is determined by the horizontal velocity and the new time of flight.
- π The final range D' is the sum of D1 and 0.9 times D1, leading to the calculation of the multiple n as 0.95.

###### Q & A

### What is the basic concept of the projectile motion problem discussed in the script?

-The script discusses a projectile motion problem where a projectile is launched at an angle Theta and speed V, with the gravitational acceleration changing at the highest point of its trajectory, affecting the range of the projectile.

### What is the formula for the range of a projectile under normal conditions?

-The range of a projectile under normal conditions is given by \( D = \frac{V^2 \sin(2\Theta)}{G} \), where V is the initial velocity, Theta is the launch angle, and G is the gravitational acceleration.

### How does the change in gravitational acceleration at the highest point affect the projectile's range?

-When the gravitational acceleration changes to \( \frac{G}{0.81} \) at the highest point, it affects the time of flight and the subsequent range of the projectile in the second half of its trajectory.

### What is the initial vertical component of the projectile's velocity?

-The initial vertical component of the projectile's velocity is \( V \sin(\Theta) \), which is used to calculate the time of flight to the maximum height.

### How is the time of flight to the maximum height calculated in the first half of the trajectory?

-The time of flight to the maximum height in the first half of the trajectory is calculated using \( T = \frac{V \sin(\Theta)}{G} \).

### What is the formula for the horizontal range (D1) in the first half of the trajectory?

-The horizontal range (D1) in the first half of the trajectory is \( D1 = V \cos(\Theta) \cdot T = \frac{V^2 \cos(\Theta) \sin(\Theta)}{G} \).

### How is the maximum height of the projectile calculated?

-The maximum height of the projectile is calculated using the equation \( V^2 = U^2 + 2aS \), where \( S = \frac{V^2 \sin^2(\Theta)}{2G} \) after rearranging for the maximum height (S).

### What is the new formula for the time of flight in the second half of the trajectory with the changed gravitational acceleration?

-The new time of flight in the second half of the trajectory is \( T' = \frac{V \sin(\Theta)}{\sqrt{0.81}G} \), taking into account the changed gravitational acceleration.

### How is the horizontal range (D2') in the second half of the trajectory calculated?

-The horizontal range (D2') in the second half of the trajectory is \( D2' = V \cos(\Theta) \cdot T' = \frac{V^2 \cos(\Theta) \sin(\Theta)}{0.9G} \).

### What is the final expression for the new range (D') of the projectile with the changed gravitational acceleration?

-The new range (D') of the projectile is \( D' = 1.9 \cdot \frac{V^2 \cos(\Theta) \sin(\Theta)}{G} \), which is 1.9 times the horizontal range in the first half of the trajectory.

### What is the value of n, which represents the multiple of the original range after the change in gravitational acceleration?

-The value of n is 0.95, which means the new range (D') is 0.95 times the original range (D) under normal conditions.

###### Outlines

##### π Projectile Motion with Variable Gravity

This paragraph discusses a physics problem involving projectile motion under altered gravitational conditions. Initially, a projectile is launched at an angle Theta and speed V, with normal gravity G, yielding a range D. However, at the projectile's highest point, gravity changes to G/0.81, affecting the range to D'. The problem requires finding the multiplier n for the new range. The solution involves calculating the time of flight and the horizontal range D1 using the initial velocity components and gravity. It then explores the conservation of energy to relate D to D1. The paragraph concludes with setting up equations for the second half of the projectile's trajectory under the new gravity condition.

##### π Calculating New Range with Altered Gravity

The second paragraph continues the projectile motion problem by focusing on the second region of the trajectory where gravity has changed. It starts by determining the maximum height reached by the projectile under the original gravity using the kinematic equation. Then, with the new gravity G' (G/0.81), it calculates the time squared T' for the projectile to fall from this height. The horizontal velocity remains constant, and the new range D2' for this region is calculated by multiplying the horizontal velocity by the new time T'. The paragraph concludes by combining the results from both regions to solve for the multiplier n, which is found to be 0.95, indicating the new range D' is 95% of the original range D.

###### Mindmap

###### Keywords

##### π‘Projectile Motion

##### π‘Gravitational Acceleration (G)

##### π‘Range (D)

##### π‘Theta (Ξ)

##### π‘Velocity Components

##### π‘Time of Flight

##### π‘Horizontal Range (D1)

##### π‘Maximum Height

##### π‘New Acceleration (G Prime)

##### π‘Energy Conservation

##### π‘Multiple (n)

###### Highlights

Solving a projectile motion problem from the JEE Advance paper with a twist - change in gravitational acceleration at the highest point.

Original range formula derived using projectile motion equations and conservation of energy.

Calculation of time of flight using vertical velocity and acceleration.

Determination of horizontal range using constant horizontal velocity.

Expression for maximum height derived using kinematic equations.

New gravitational acceleration leads to a new time of flight calculation.

Horizontal velocity remains constant despite the change in acceleration.

Second half range calculation using the new time of flight and horizontal velocity.

Comparison of the first and second half ranges to find the new range as a multiple of the original.

Derivation of the new range formula using the derived expressions for D1 and D2'.

Simplification of the new range formula by canceling common factors.

Final calculation of the multiple n, which is the ratio of the new range to the original range.

Result of n being 0.95, indicating the new range is 95% of the original range.

The problem tests understanding of projectile motion, kinematic equations, and conservation of energy.

The solution process involves multiple steps of derivation and calculation.

The problem requires analytical skills to handle the change in acceleration and its impact on range.

The solution provides a clear, step-by-step approach to solving the complex projectile motion problem.

###### Transcripts

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