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In the below question, what does the phrase "the time taken by the particle to hit the ground" mean? Does it mean time taken by the ball to cover distance from the point of projection to the ground or does it mean the time taken by the ball to cover the distance from the point of projection when it is under free fall?

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is

a) 2gH = n²u²

b) gH = (n-2)²u²

c) 2gH = nu²(n-2)

d) gH = (n-2)²u²

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  • $\begingroup$ Hint: Symmetrically change in velocity from $0-u$ and than $u-0$. So, the time taken will be twice as the first one. $$⟹n=2$$ $\endgroup$
    – user354639
    Commented Jan 5, 2023 at 19:21
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    $\begingroup$ To me it reads like the ball goes upward some distance $d$ and then falls a distance $H+d$ to the ground. Otherwise there is no point in specifying a tower. $\endgroup$
    – RC_23
    Commented Jan 6, 2023 at 4:17

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The time taken to reach highest point is,

$$t_1=\sqrt{\frac{2H}{g}}=\frac{u}{g} $$

And when the particle reach the bottom again, displacement in y is 0. So,

$$0=ut_2-\frac{gt_2^2}{2}⟹t_2=\frac{2u}{g}$$

So, $2t_1=t_2⟹n=2$. Now you can see according to your options.

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