Menu Close

A-balloon-ascends-vertically-with-a-constant-speed-for-5-seconds-when-a-pebble-falls-from-it-reaching-the-ground-in-5-s-The-speed-of-the-balloon-is-




Question Number 15116 by Tinkutara last updated on 07/Jun/17
A balloon ascends vertically with a  constant speed for 5 seconds, when a  pebble falls from it reaching the ground  in 5 s. The speed of the balloon is?
Aballoonascendsverticallywithaconstantspeedfor5seconds,whenapebblefallsfromitreachingthegroundin5s.Thespeedoftheballoonis?
Answered by ajfour last updated on 07/Jun/17
pebble′s initial velocity = velocity  of balloon=u  s=ut+(1/2)at^2   displacement of pebble by the  time it reaches ground = height  througb which balloon ascends  in 5 seconds.  −ut_0 =ut−(1/2)gt^2   t_0 =5s and  t=5s , so  −5u=5u−(1/2)(10)(25)   ⇒ 10u =125   u=12.5 m/s .
pebblesinitialvelocity=velocityofballoon=us=ut+12at2displacementofpebblebythetimeitreachesground=heightthrougbwhichballoonascendsin5seconds.ut0=ut12gt2t0=5sandt=5s,so5u=5u12(10)(25)10u=125u=12.5m/s.
Commented by Tinkutara last updated on 07/Jun/17
Thanks Sir!
ThanksSir!
Answered by mrW1 last updated on 07/Jun/17
v5=−v5+(1/2)×10×5^2   v=12.5 m/s
v5=v5+12×10×52v=12.5m/s
Commented by Tinkutara last updated on 07/Jun/17
Thanks Sir!
ThanksSir!

Leave a Reply

Your email address will not be published. Required fields are marked *