Two particles released from the same height and reached the base at the same time

Problem
Particles A and B are elevated 12 meters high from a given reference base. Particle A is projected down an incline of length 20 meters at the same time particle B is let to freely fall vertically. Find the velocity of projection of particle A if both particles strikes the base at the same time.

Solution

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For particle B

$h = \frac{1}{2}gt^2$

$12 = \frac{1}{2}gt^2$

$t = \sqrt{\dfrac{24}{g}}$

For particle A

$\dfrac{v_{Ai-y}}{v_{Ai}} = \dfrac{12}{20}$

$v_{Ai-y} = 0.6v_{Ai}$

Apply the formula s = v_ot + 0.5at² to the vertical movement of A
$s = v_ot + \frac{1}{2}at^2$

$h = v_{Ai-y}\,t + \frac{1}{2}gt^2$

$12 = (0.6v_{Ai})\sqrt{\dfrac{24}{g}} + \dfrac{g}{2}\left( \sqrt{\dfrac{24}{g}} \right)^2$

$12 = 0.6v_{Ai} \, \sqrt{\dfrac{24}{g}} + \dfrac{g}{2}\left( \dfrac{24}{g} \right)$

$12 = 0.6v_{Ai} \, \sqrt{\dfrac{24}{g}} + 12$

$v_{Ai} = 0$ answer

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