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
for the first particle final velocity 0 is achieved in ..
v = u - gt
t = -30 / -10 = 3 seconds ------------------------------------------ (i)
distance travelled
S = ut - gt^2 / 2
S = 90 - 90/2
S = 90 - 45 = 45 m
assume that the second body is thrown after 'x' seconds.
it has to reach the top at the same time as that of the 1st particle because of the velocity but at some point they will collide.
they collide at a point when the first particle is returning.
distance travelled in 4 seconds = 45 + ut + at^2 / 2
= 45 + 0 + 5 = 50m but the displacement is 50-10 = 40m from the ground
distance travelled by second body in 2 seconds = ut - at^2/2
= 60 - 10*4/2
= 60 - 20 = 40m
so the two particles collide at a height of 40m when the first body has travelled for 4 seconds while the second body has travelled for 2 seconds, so the time difference gives the 'T' (time after the throw of 1st body)
T = 4-2 = 2 seconds
so the second body was thrown 2 seconds after the first body was thrown.
-----------------------------------------------------------------------------------------------------------
PLEASE APPROVE
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 !