y = (P t^2 - Q t^3) m is the vertical displa | vedclass

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(C) Velocity is defined as the rate of change of displacement.$v_y = \frac{dy}{dt}$$= \frac{d(P t^2 - Q t^3)}{dt}$$= 2Pt - 3Qt^2$At maximum height,the

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$\frac{4 P^3}{27 Q^2}$

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(C) Velocity is defined as the rate of change of displacement.$v_y = \frac{dy}{dt}$$= \frac{d(P t^2 - Q t^3)}{dt}$$= 2Pt - 3Qt^2$At maximum height,the vertical velocity of the particle becomes zero.$2Pt - 3Qt^2 = 0$$t(2P - 3Qt) = 0$Since $t=0$ is the start of the motion,the time at maximum height is $t = \frac{2P}{3Q}$.Now,substituting this time into the displacement equation:$y_{max} = P(\frac{2P}{3Q})^2 - Q(\frac{2P}{3Q})^3$$= P(\frac{4P^2}{9Q^2}) - Q(\frac{8P^3}{27Q^3})$$= \frac{4P^3}{9Q^2} - \frac{8P^3}{27Q^2}$$= \frac{12P^3 - 8P^3}{27Q^2}$$= \frac{4P^3}{27Q^2}$

$y = (P t^2 - Q t^3) \ m$ is the vertical displacement of a ball which is moving in a vertical plane. Then the maximum height that the ball can reach is,

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