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
Let Ub and Vb be the velocities at t=0 and t=2 for the balloon
And accelaration of balloon(Ab)=4.9=g/2 m/s2
Vb=Ub+(Ab)t (Ub=0)
from this we get, Vb=g m/s (Upwards)
and from ,s=(Ub)t+(1/2)(Ab)t^2 (where s is the hieght at which the ball is dropped)
we get s=g meters
when the ball is dropped its initial velocity is gm/s upwards(due to balloon)
hence velocity of ball at ground(using third eq of motion):
V^2=g^2 + 2(g)(g) (since,,initial vel of ball=g m/s,,,acc due to gravity=g m/s2,,,displacement=g meters)
v^2=3g^2------------------------------------1
If we want to find the greatest hieght(to which it rises)then it should rebound back with same velocity(elastic collision)
therfore ,
v^2=u^2-2(g)(H)---------------------------2 (Where H is the max hieght it reaches)
in this case v=0 since at the greatest hieght the vel of ball is zero
and u^2=3g^2 (from eq--1)
therefore
eq 2 becomes:
3g^2=2gs
ie,s=3/2(g)=14.7m
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 !