Question

The equation of path of a projectile moving in x-y plane is given by $ y = x - \frac{x^2}{25} $, then the initial speed and maximum height of particle will be (take $ g = 10 \, \text{m/s}^2 $):

• A. $ 4\sqrt{10} \, \text{m/s} $ and $ \frac{29}{4} \, \text{m} $
• B. $ 25\sqrt{10} \, \text{m/s} $ and $ \frac{21}{4} \, \text{m} $
• C. $ 2\sqrt{10} \, \text{m/s} $ and $ \frac{31}{4} \, \text{m} $
• D. $ 5\sqrt{10} \, \text{m/s} $ and $ \frac{25}{4} \, \text{m} $

Hint:

To find the maximum height quickly from a trajectory equation $ y = ax - bx^2 $, you can use calculus. The maximum height occurs at the vertex where $ dy/dx = 0 $. Here, $ 1 - (2x/25) = 0 \implies x = 12.5 $. Plugging $ x = 12.5 $ into the equation gives $ y = 12.5 - (12.5)^2/25 = 12.5 - 6.25 = 6.25 $, which is exactly $ 25/4 $.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Compare the given trajectory equation with the standard projectile motion equation \( y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta} \) to find angle \( \theta \) and initial velocity \( u \).
Step 2: Detailed Explanation:
1. \( \tan \theta = 1 \implies \theta = 45^\circ \).
2. \( \frac{g}{2u^2 \cos^2 45^\circ} = \frac{1}{25} \implies \frac{10}{2u^2(1/2)} = \frac{1}{25} \implies u^2 = 250 \implies u = 5\sqrt{10} \, \text{m/s} \).
3. Max Height \( H = \frac{u^2 \sin^2 \theta}{2g} = \frac{250 \times (1/2)}{20} = \frac{125}{20} = \frac{25}{4} \, \text{m} \).
Step 3: Final Answer:
Initial speed is \( 5\sqrt{10} \, \text{m/s} \) and max height is \( \frac{25}{4} \, \text{m} \).