Question

A body thrown vertically upwards attains a
maximum height H. While moving
upwards if it covers first 3H/4 distance
from the ground in time T, then time taken
to cover remaining distance H/4 will be? ​

Answer 1

Explanation:

Hey A2A

Let P be the point at which Height =3H43H4

u= initial velocity of projectile

Total Time(T)=ug(1)(1)(T)=ug

Total Height (H)=u22g(H)=u22g

Or H=T2g2(2)(2)H=T2g2

Equations for point P

So, v2p=u2−2g(3H)4(3)(3)vp2=u2−2g(3H)4

3H4=utp−gt2p2(4)(4)3H4=utp−gtp22

ON solving this (4th) quadratic equation and putting values of uu from 1 & HH from 2

You get 2 values of tptp namely

tp=3T2andtp=3T2and

tp=T2tp=T2

1st value is possible only after the projectile is on its return path as tp>Ttp>T

HENCE taking second value of tptp

Time taken to reach P =tp=T2

So time taken to cover remaining height = T-T2=T/2T2=T/2

Fascinating it is that it takes T2T2units to cover 3H43H4

So it would take equal time from 0 to 3H43H4 and 3H43H4 to HH

( I have considered the time from 00 to HH as T.)T.)

But here in this problem time for 3H43H4 is given as T.T.

But the last result is independent of any such thing.

So it would always take equal time from 0 to 3H43H4 and 3H43H4 to HH

Answer 2

Answer:

To cover 3H/4 height it takes T time

So using Unitary method;

3H/4 =T

H/4=T/3

Explanation: