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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 time that taken by it to reach the highest to its path. Show that 2gH =nu²(n-2)

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 time that taken by it to reach the highest to its path. Show that 2gH =nu²(n-2)

Grade:11

3 Answers

Soumya Ranjan Mohanty
117 Points
6 years ago
Dear student,We know time to reach maximum height t1=u/g. t2 is time to reach ground is given by equation:-h=ut2 - 1/2 * g*t^2Given that t2=nt1 , putting value of t2 in above equation, we get 2gh=nu^2(n-2).Thanking you, Soumya Ranjan Mohanty.
tushar jaryal
38 Points
6 years ago
As H is the height of tower
time taken by particle to reach highest point of its path is t=u/g
so speed on reaching ground,v=2+2gH>^2
now v=u + at
=>v= -u + gt’
=>2+2gH>^2= -u + gt[from above]
on solving
t’=[u+2+2gH>^2]/g 
given that time taken by particle to hit ground is n times that taken by it to reach the highest point.
=>t’=[u+2+2gH>^2]/g =nu/g
on solving
2gH=n[n-2]u2
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the attached image of the solution to your problem.

Thanks and Regards
645-659_5.PNG

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