Question

The trajectory of projectile, projected from the ground is given by y=x-x²/20. Where x and y are measured in meter. The maximum height attained by the projectile will be

• A. 5m
• B. 10√2m
• C. 10m
• D. 200m

Answer 1

The Correct Option is A

The trajectory of the projectile is given by: \[ y = x - \frac{x^2}{20}. \] The maximum height corresponds to the point where the slope of the trajectory (\( \frac{dy}{dx} \)) is zero. 
Step 1: Differentiate \( y \) with respect to \( x \). \[ \frac{dy}{dx} = 1 - \frac{2x}{20} = 1 - \frac{x}{10}. \] At the maximum height: \[ \frac{dy}{dx} = 0 \implies 1 - \frac{x}{10} = 0 \implies x = 10 \, \text{m}. \] 
Step 2: Calculate the maximum height. Substitute \( x = 10 \) into the equation for \( y \): \[ y = 10 - \frac{10^2}{20}. \] Simplify: \[ y = 10 - \frac{100}{20} = 10 - 5 = 5 \, \text{m}. \] 
Final Answer: The maximum height attained by the projectile is: \[ \boxed{5 \, \text{m}}. \]