
ntAB=tAB+tBC+tCD
(n−1)tAB=tBC+tCD....(1)
(n−1)gu=gu+tCD
∴(n−2)gu=tCD ...(2)
H+h=21gtBD2
H+2gu2=21g[g(n−1)u]2 ...from (1)
∴H+2gu2=2g(n−1)2u2
∴H=2gu2[(n−1)2−1]
2g H = nu2(n - 2)
JEE Main 2014 — Physics Mechanics
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 :
Held on 6 Apr 2014 · Verified 6 Jul 2026.
2 g H = n2u2
g H = (n - 2)2u2
2g H = nu2(n - 2)
g H = (n - 2)u2
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