Two particles are projected vertically upwards with the same velocity on two different planets with accelerations due to gravities $g_{1}$ and $g_{2}$ respectively. If they fall back to their initial points of projection after lapse of times $t_{1}$ and $t_{2}$ , respectively, then

t_{1} t_{2} = g_{1} g_{2}

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t_{1} g_{1} = t_{2} g_{2}

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t_{1} g_{2} = t_{2} g_{1}

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{t_{1}}^{2} + {t_{2}}^{2} = g_{1} + g_{2}

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Solution

The correct option is A $t_{1} g_{1} = t_{2} g_{2}$
For both particles, net vertical displacement is zero.
Particle 1: $y = u t + \frac{1}{2} a t^{2} \Rightarrow 0 = u t_{1} - \frac{1}{2} g_{1} t_{1}^{2} o r u = \frac{1}{2} g_{1} t_{1}$
Particle 2: $0 = u t_{2} - \frac{1}{2} g_{2} t_{2}^{2} o r u = \frac{1}{2} g_{2} t_{2}$
thus $u = {\frac{1}{2} g}_{1} t_{1} = \frac{1}{2} g_{2} t_{2} o r g_{1} t_{1} = g_{2} t_{2}$

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