Question

If the equation of motion of a projectile is \( Y = AX - B X^2 \), then the ratio of the maximum height reached and the range of the projectile is:

• A. \( \frac{A}{4} \)
• B. \( \frac{A}{B} \)
• C. \( \frac{B}{4} \)
• D. \( \frac{A^2}{B} \)

Hint:

For projectile equations in the form \( Y = AX - B X^2 \), use: \[ H_{\max} = \frac{A^2}{4B}, R = \frac{A}{B} \] to compute the height-to-range ratio efficiently.

Answer 1

The Correct Option is A

The given equation of motion: \[ Y = AX - B X^2 \] Step 1: Finding Maximum Height At the peak of the trajectory, \( \frac{dY}{dX} = 0 \). Differentiating: \[ \frac{dY}{dX} = A - 2BX \] Setting \( \frac{dY}{dX} = 0 \), \[ A - 2BX = 0 \] Solving for \( X \), \[ X = \frac{A}{2B} \] Substituting into the equation of motion: \[ H_{\max} = A \cdot \frac{A}{2B} - B \left( \frac{A}{2B} \right)^2 \] \[ H_{\max} = \frac{A^2}{2B} - \frac{A^2}{4B} = \frac{A^2}{4B} \] Step 2: Finding Range The projectile reaches the ground when \( Y = 0 \). Setting \( Y = 0 \): \[ AX - B X^2 = 0 \] \[ X(A - BX) = 0 \] Solving for \( X \), \[ X = \frac{A}{B} \] Thus, the range \( R \) is: \[ R = \frac{A}{B} \] Step 3: Computing the Ratio \[ \frac{H_{\max}}{R} = \frac{\frac{A^2}{4B}}{\frac{A}{B}} \] \[ = \frac{A}{4} \] Thus, the correct answer is: \[ \frac{A}{4} \]