Join now for JEE/NEET and also prepare for Boards Learn Science & Maths Concepts for JEE, NEET, CBSE @ Rs. 99! Register Now
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
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
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
its indeed 20.32......how did u get it?plz post it.....plz...plz...
gosh dats cool i got it right yeah.........
I would surely tell you....
C assume the initial speed to be 'u'
we got a distace of 8 m to travel before the net which is 2.24 m high. since the original height at which the ball was hit was 3 m, so the difference in height is 3-2.24=.76m. Using horizontal projectile formula i.e. range = u(2h/g)1/2
we get 8=u[(2 x .76)/9.8]1/2
: u=8[9.8/(2 x .76)]1/2
: u=20.32m/s.
Dats all.
Hey I am from Shimla atleast got smeone to interact wid who talks about studies. Could u give me ur e-mail ID ill keep sendin you sme questions.
IT would be kind of you to send me your E-mail ID soon enough. Nd where r u from which skul n which class......
I have been longing to interact with smeone on this site thanks for considering my answer. Could you please give me ur email ID
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 !