A stone is dropped from height h. Simultaneously, another stone is thrown up from the ground which reaches a height 4h. The two stones will cross each other after time:-
Answers
Answer:
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Explanation:
Let the initial velocity of stone which is thrown from the ground u
2
.
At maximum height 4h, Velocity v
2
=0
Using v
2
=u
2
−2as ⇒0=u
2
2
−2g×4h ⇒u
2
=
8gh
...1
Let h
1
be the distance covered by the stone which is dropped from height h,
Let h
2
be the distance covered by the stone which is thrown up from the ground.
Let at time t both stone crossed each other.
Using s=ut+
2
1
at
2
,
⇒−h
1
=−
2
1
gt
2
⇒h
1
=
2
1
gt
2
....2
⇒h
2
=u
2
t−
2
1
gt
2
....3
Adding h
1
and h
2
gives the height h.
∴ Adding equations 2 and 3,
⇒h=
2
1
gt
2
+u
2
t−
2
1
gt
2
⇒h=u
2
t
Using the value of u
2
=
8gh
,
⇒h=
8gh
t ⇒t=
8g
h
Answer:Let the initial velocity of stone which is thrown from the ground u
2
.
At maximum height 4h, Velocity v
2
=0
Using v
2
=u
2
−2as ⇒0=u
2
2
−2g×4h ⇒u
2
=
8gh
...1
Let h
1
be the distance covered by the stone which is dropped from height h,
Let h
2
be the distance covered by the stone which is thrown up from the ground.
Let at time t both stone crossed each other.
Using s=ut+
2
1
at
2
,
⇒−h
1
=−
2
1
gt
2
⇒h
1
=
2
1
gt
2
....2
⇒h
2
=u
2
t−
2
1
gt
2
....3
Adding h
1
and h
2
gives the height h.
∴ Adding equations 2 and 3,
⇒h=
2
1
gt
2
+u
2
t−
2
1
gt
2
⇒h=u
2
t
Using the value of u
2
=
8gh
,
⇒h=
8gh
t ⇒t=
8g
h
Explanation: