Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
A body is thrown up wuth initial velocity 'u' .and reaches height 'h'
To reach double height it must be projected with initial velocity of-
if the particle is projected with velocity u then at maximum height its velocity becomes 0
we have ,
V2 = U2 - 2gHmax
U2 = 2gHmax or
U2 directly proportional to Hmax
(Ui)2 / (Uf)2 = (Hmax)i/(Hmax)f
Ui = u , Uf=? , (Hmax)i = h , (Hmax)f = 2h
after substituting these
Uf = (sqrt2)u ans
the height attained by the body is h therefore applying third equation of motion
v^2=u^2-2gh
at height h final velocity is zero
u^2=2gh
u=square root(2gh) ....1
now height attained by the body is 2h
therefore putting 2h in place of h
the equation comes to be
u'^2=2g(2h)
u'= square root(2*2gh)
but square root 2gh=u
therefore u'=u*square root(2)
it is given that 'h' height is attained by a body and its velocity is u...
writing eqn of motion...
v(sqr)-u(sqr)=2as
putting values....
v=0,u=u,and a=-g(coz g is in downward direction), s=h
-u(sqr)=2(-g)h
u(sqr)=2gh
so to get height of 2h u should be sqrroot(2) to satisfy above eqn
answer is sqrroot(2)
root 2 times the original velocity
It is sqrt 2 times u.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !