Question

A ball follows the vertical displacement \( y = (Pt^2 - Qt^3) \), where \( y \) is in meters. Find the maximum height the ball can reach.

• A. \( \frac{27P^3}{4Q^2} \)
• B. \( \frac{4Q^2}{27P^3} \)
• C. \( \frac{4P^3}{27Q^2} \)
• D. \( \frac{27Q^2}{4P^2} \)

Hint:

To find maximum of a displacement-time function, differentiate and set derivative to zero.

Answer 1

The Correct Option is A

Maximum height occurs when \( \frac{dy}{dt} = 0 \): \[ \frac{dy}{dt} = 2Pt - 3Qt^2 = 0 \Rightarrow t = \frac{2P}{3Q} \] Substitute in y: \[ y = P\left(\frac{4P^2}{9Q^2}\right) - Q\left(\frac{8P^3}{27Q^3}\right) = \frac{4P^3}{9Q^2} - \frac{8P^3}{27Q^2} = \frac{12P^3 - 8P^3}{27Q^2} = \frac{4P^3}{27Q^2} \] Correction: Answer should be (3) \( \frac{4P^3}{27Q^2} \) — mismatch in answer key noted.