If a body is thrown vertically up with a velocity u. It passes a point at a height h above the ground at time t1 while going up and at time t2 while falling down. Then the relation between t1, t2 and h is:

Here equation for displacement is h=ut-12gt2 t2-2utg+2hg=0Product of roots is t1t2=2hg

If a body is thrown vertically up with a velocity u. It passes a point at a height h above the ground at time t1 while going up and at time t2 while falling down. Then the relation between t1, t2 and h is:

Here equation for displacement is h=ut-12gt2 t2-2utg+2hg=0Product of roots is t1t2=2hg

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If a body is thrown vertically up with a velocity $u$ . It passes a point at a height $h$ above the ground at time t₁ while going up and at time t₂ while falling down. Then the relation between t₁, t₂ and $h$ is:

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$t_{1} t_{2} = \frac{2 h}{g}$

$t_{1} t_{2} = \frac{h}{g}$

${(t_{1} + t_{2})}^{2} = \frac{2 h}{g}$

${(t_{1} + t_{2})}^{2} = \frac{h}{g}$

answer is A.

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$Here equation for displacement is h = ut - \frac{1}{2} {gt}^{2} t^{2} - \frac{2 ut}{g} + \frac{2 h}{g} = 0$

Product of roots is $t_{1} t_{2} = \frac{2 h}{g}$

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If a body is thrown vertically up with a velocity u. It passes a point at a height h above the ground at time t1 while going up and at time t2 while falling down. Then the relation between t1, t2 and h is:

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If a body is thrown vertically up with a velocity $u$ . It passes a point at a height $h$ above the ground at time t₁ while going up and at time t₂ while falling down. Then the relation between t₁, t₂ and $h$ is: