IIT JEE , AIEEE Forum - Physics , Mechanics

sgyaan

1.a balloon rises 4rm d ground wid a constant accln of 3m/s².5 sec later,a stone is thrown up 4rm d launchin site of d balloon.Wat must be d min vel of d stone so dat it just touches d balloon? {g=10m/s²& sqrt(5.77)=2.4}

i just need 2 knw d method coz I m nt getting d ans wid mine?..

hash_include

equate the heights. and then use t2 - t1 = 5 sec
where t2 = time balloon takes to go to that height and t1 = time stone takes

ganesha1991

edit

sgyaan

c we hv 3 unknowns wid ur method,u,t1 & t2...wat abt d 3rd eqn???

karthik2007

I'll give you the algorithm:

First calculate the distance traversed by the balloon in 5 seconds. Now, the stone is thrown. Let the stone meet the balloon at some t seconds after its thrown. Calculate the distance S traversed by the balloon in this time t as a function of t. The stone will travel a distance = S + distance travelled by balloon in 1st five seconds in the same time t. Also use the fact that final velocity of the stone is zero.

sgyaan

karthik dis ws d same method i used,but i wsn't gettin d ans....

in d soln it ws written dat at d time they both meet each other ,dere velocities wil be equal........i wsn't gettin hw?

ganesha1991

edit

ganesha1991

plz tell the sol. i'll try to solve it

sgyaan

ok, d ans is approx 46m/s

hash_include

distance travelled by balloon in first 5 sec = 1/2 * 3 * 25 = 75/2

vel of balloon after 5 sec = 3*5 = 15

let the balloon and stone meet (t + 5) sec after the balloon starts going up

h = ht at which they meet = 1/2gt^2

also, h1 = distance traveled by balloon from T=5sec till T=t+5
h1 = 15*t + 1/2*3*t^2

h - h1 = 75/2
solve for t and get hence get u from 0 = u - gt

ganesha1991

hash-include
i followed the same thing
but not getting the answer

hash_include

me 2.. but i found out one thing..
if the qsn is "After the balloon travels 5 metres, the stone is thrown", the answer will come 46m/s properly..

madman

there are two things here
the velocities should be equal when they meet , so that it just touches and the distances travelled by them should be equal
velocity of the balloon after 5s = 3t=15 m/s
let them meet after time t
then
velocity of balloon after ts = 15+3t
velocity of stone= u - gt
u - gt = 15 + 3t -1
or
ut - gt^2 = 15t + 3t^2 -2
distance travelled by stone in first 5s=1/2 * 3 *25 = 75/2
distance travelled in the next ts = 15t + 0.5*3t^2
totla distance travelled = 75/2 + 15 t + 3/2 t^2 -3
distance travelled by stone = ut - 1/2 g t^2 -4
ut - 1/2 g t^2 = 75/2 + 15 t + 3/2 t^2
therefore
1/2 g t^2 = 75/2 -3/2 t^2
10t^2 = 75 -3t^2
t^2 = 75/13 = 5.77
t = 2.4
v = 24 + 15 + 7.2 = 46.2 m/s

sgyaan

pls madman i do need an explanation as 2 why velocities wil be equal wen they meet..

i think dat 4 d min vel of d stone it's vel at d tym of touchin d balloon shld be 0?

hash_include

ur proof is wrong madman.. pls check carefully
you've made quite a few mistakes..

Ask & Discuss Questions with Community & Experts