Sponsor Area

Motion In Straight Line

Question
CBSEENPH11020269

From the 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)2u2

  • gH = (n-2)2u2

Solution

C.

2gH = nu2(n-2)2u2

Time is taken to reach the maximum height t1 = u/g
If t2 is the time taken to hit the ground,
i.e, negative straight H space equals space ut subscript 2 minus 1 half gt subscript 2 superscript 2
straight t subscript 2 space equals space nt subscript 1
minus straight H space equals space straight u nu over straight g minus 1 half straight g fraction numerator straight n squared straight u squared over denominator straight g squared end fraction
minus straight H space equals space nu squared over straight g minus 1 half fraction numerator straight n squared straight u squared over denominator straight g end fraction
2 gH space equals nu squared left parenthesis straight n minus 2 right parenthesis