JEE Mains 2025 Physics PYQ | Projectile Motion | 8 April Shift 2
This post covers a JEE Mains 2025 Physics Previous Year Question (PYQ) from the 8th April Shift 2 paper. The topic is Projectile Motion, and the problem tests your understanding of the relationship between maximum height and time of flight. We provide a clear step-by-step solution along with a full video explanation.
Question
Projectile Motion JEE Mains 2025 Projectile Motion – 8th April Shift 2
Two balls with same mass and initial velocity, are projected at different angles in such a way that the maximum height reached by the first ball is 8 times higher than that of the second ball. \( T_1 \) and \( T_2 \) are the total flying times of the first and second ball, respectively. Then the ratio of \( T_1 \) and \( T_2 \) is:
- 2√2 : 1
- √2 : 1
- 4 : 1
- 2 : 1
Detailed Step-by-Step Solution
Step 1: Use the formula for maximum height
The formula for maximum height in projectile motion is:
\( H = \frac{u^2 \sin^2 \theta}{2g} \)
Given:
\( H_1 = 8H_2 \)
So,
\[
\frac{H_1}{H_2} = \frac{\sin^2 \theta_1}{\sin^2 \theta_2} = 8
\Rightarrow \left( \frac{\sin \theta_1}{\sin \theta_2} \right)^2 = 8
\Rightarrow \frac{\sin \theta_1}{\sin \theta_2} = 2\sqrt{2}
\]
Step 2: Use the formula for time of flight
Time of flight formula:
\[
T = \frac{2u \sin \theta}{g}
\]
So the ratio of total flight times becomes:
\[
\frac{T_1}{T_2} = \frac{\sin \theta_1}{\sin \theta_2} = 2\sqrt{2}
\]
✅ Final Answer:
Option (1): 2√2 : 1
Watch the Full Video Solution
Watch the complete solution with explanation on our YouTube channel:
Conclusion
This question demonstrates how comparing height and time formulas in projectile motion leads to quick results. Understanding trigonometric ratios and their influence on motion is key for mastering kinematics.
