51. Physics | Kinematics | Distance Between Dropped Stones | by Ashish Arora

51. Physics | Kinematics | Distance Between Dropped Stones | by Ashish Arora


in this illustration, we’ll discuss about
the distance between, dropped stones. and the situation is a balloon is ascending upward
with acceleration 1 meter per second square. and 2 stones are dropped from it at an interval
of 2 seconds. and we are required to find the distance between the stones 1 point 5
second after the second stone is released. here, to analyze the situation better we,
draw a picture. say this is, a balloon which is going up at a speed u. and we also given
that it is going up with an acceleration ay is 1 meter per second square. here if, a stone.
1 is dropped the stone 1 will also be having a speed u. so it’ll go up and then it’ll
come down. so here we can say after. 3 point 5 second, because initially after 2 second,
the second stone is dropped and we are required to find the distance at 1 point after 5 second.
say stone 1 is at a height of, h 1. from the initial point from where the stone 1 was dropped
it’ll go and it’ll come down. and it’ll be retarded by g this g is the retardation
in stone. or it is having a downward acceleration g. so here we can calculate the displacement.
of stone 1. after 3 point 5 second is. this can be written as h 1 which is u t. plus,
half ay t square, so here we can write. this is u multiplied by 3 point 5. and ay can be
taken as minus g so this minus. half, g can be taken as 10 multiplied by 3 point 5, whole
square. so here the value of h 1, is. 3 point 5 u, minus. this is, 6 1, point 2 5. this
is the displacement of stone 1. from t equal to zero at time t is equal to 3 point 5 second.
and in 2 second the balloon will also ascend by some distance. and if we calculate the
distance which it’ll ascend is say delta h. so if this is the location of balloon.
at t equal to 2 second from where the second stone is dropped. so here we can see. the
displacement of balloon. in 2 second is, this can be written as delta h which is u multiplied
by 2 plus half. ay t square ay we can take as 1 because we are given that balloon is
accelerating with 1 meter per second square. multiplied by 2 square so this is 2 u plus.
2. this is the displacement of balloon in 2 second. from here. the stone 2 is released
so it will also have an upward velocity v which. is their in the balloon. so, this velocity
v also we can calculate. velocity of balloon. after 2 second is, this v can be written as
u plus 80. so this is u plus. ay is 1 and t is 2 second so this is u plus 2. meter per
second. now with this speed stone 2 is released it will go up and then finally it will come
down. so in this situation say here, it is. at a position h 2 from the initial, location.
from where it is drop so, if this is h 2. here we can calculate. the displacement. of
stone 2. in 1 point 5 second i am taking 1 point 5 second because. this is again the
time t is 3 point 5 second where we have calculated, the location of stone 1. and at this instant
we are required to find. the separation between the 2 stones. so this displacement h 2 can
be calculated as, this initial speed is u plus 2 multiplied by 1 point 5, minus half.
acceleration is g which can be taken as 10 multiplied by 1 point 5, whole square. so
on simplifying this we are getting it as 1 point 5 u. plus 3. minus, here this can be
written as, 11 point, 2 5 numerically i have solved it. so in this situation. if we, wish
to, find out the separation between the 2 stones here we can calculate. that is separation
between. stones. at t is equal to 3 point 5 second, this can be given as. s and this
will be, h 2 plus delta h. minus h 1, that will be the separation between the 2 stones.
this is s. this delta h plus h 2 minus h 1. so if i substitute the values. here the value
of h 2 is, this we have calculated. 1 point 5 u plus 3 minus. 11 point 2 5 plus delta
h here as, we have calculated as. 2 u plus 2. minus the value of h 1 we have, calculated
as 3 point 5 u minus, 61 point, 2 5. so further if we solve it, you can see. this 3 point
5 u 2 u will be minus 1 point 5 this gets cancelled out. so numerically, this 3 plus
2 becomes 5. and the 61 point 2 5. minus 11 point 2 5 becomes. this is plus 50 so the
result is 55 meter that will be the answer to this problem.

9 comments / Add your comment below

  1. sir why u have taken h1 and h2 positive .. it might have been negative??
    please tell me where i am getting wrong

  2. Why are we not considering the possibility that h1 can be below the point of release ? In that case the distance won't be given by the expression ( h2 + ∆h – h1 )

  3. sir can we do this question by assuming initial velocity of the stones 0 then applying 2nd equation of motion and in place of g putting value -1-10 = -11

  4. Sir you have taken velocity of stone in ground frame = velocity of balloon bcoz it is dropped wrt balloon.
    But why did not acceleration taken g-1 ??

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