JEE MAIN - Physics (2014 (Offline) - No. 29)
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:
2gH = n2u2
gH = (n - 2)2u2
2gH = nu2(n - 2)
gH = (n - 2)u2
Explanation
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Time taken to reach highest point is $$t = {u \over g}$$
Time taken by the particle to reach the ground = $$nt = {nu \over g}$$
Speed on reaching ground $$v = \sqrt {{u^2} + 2gH} $$
Now, $$v = u + at$$
$$ \Rightarrow \sqrt {{u^2} + 2gH} = - u + gt$$
$$ \Rightarrow t = {{u + \sqrt {{u^2} + 2gH} } \over g} = {{nu} \over g}$$ (from question)
$$ \Rightarrow 2gH = n\left( {n - 2} \right){u^2}$$
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