A particle is dropped from height 100 metre and particle is projected vertically up with velocity 50 metre per square from the ground around the same line find out the position where the particle will reach. (take g= 10m/s) Ans- 80 m above the ground.
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Answered by
23
method 1 :-
let at t time particle approach to each other .
1st particle which is dropped from 100 m take distance(S1) = 1/2(10)t²
and second particle which is thrown with ground take distance at t time
S2= 50t -1/2(10)t²
now,
S1 + S2 = 100m
5t² + 50t -5t² = 100
t = 2sec
now,
distance from ground they approach to each other = 50t -5t² = 100 -20 = 80 m
method 2 :-
you can solve this easily with help of relative concept
use ,
t = relative displacement /relative velocity
= (100-0)/( 50+0) = 2 sec
so, distance covered 2nd particle in 2sec = ut +1/2at² = 50t -5t² = 80 m
let at t time particle approach to each other .
1st particle which is dropped from 100 m take distance(S1) = 1/2(10)t²
and second particle which is thrown with ground take distance at t time
S2= 50t -1/2(10)t²
now,
S1 + S2 = 100m
5t² + 50t -5t² = 100
t = 2sec
now,
distance from ground they approach to each other = 50t -5t² = 100 -20 = 80 m
method 2 :-
you can solve this easily with help of relative concept
use ,
t = relative displacement /relative velocity
= (100-0)/( 50+0) = 2 sec
so, distance covered 2nd particle in 2sec = ut +1/2at² = 50t -5t² = 80 m
plz ans this one
Answered by
2
Answer:80m
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