A particle is projected vertically up and another is let fall to meet at the same instant.if they have same velocities equal in magnitude when they meet ,the distances travelled by them are in the ratios
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Answers
Answer:
Explanation:
Before we solve the question
We need to know 2 formulas
V²-U²=2as
V=U+at
In case 1 :distance travelled by ball left to fall
V=U+at
Final velocity=V
Initial velocity=0
acceleration=g
time taken
Hence V=g*t
Note: Time taken by both balls to meet is same
Using V²-U²=2as
g²t²-0=2gs
gt²/2=S
In case 2: When ball is projected vertically up
Final velocity=V
Initial velocity=U
Time taken=t
acceleration=g
Applying formula V=U+at
V=U- g*t
As finally velocity obtained in case 1 and 2 is same
gt=U-gt
U=2gt
Applying formula V²-U²=2as
g²t²-4g²t²=-2gs
-3g²t²=-2gs
3gt²/2=S
Ratio of distances travelled
Case1/Case2 =3/4
Hope it helps
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Answer:
Let A &B meet at P.
For A⇒v=gt..................(1)
For B⇒v
′
=u−gt....................(2)
The magnitude of v and v
′
to be same
∣v∣=∣v
′
∣
⇒u−gt=gt
u=2gt⇒t=
2g
u
So, dist. travelled by B:
⇒S=ut−
2
1
gt
2
S=u(
2g
u
)−
2
1
g(
2g
u
)
2
S=
2g
u
2
−
8g
u
2
=
8
3
(
g
u
2
)
Dist. travelled by B⇒S
′
=
2
1
gt
2
=
2
1
g(
2g
u
)
2
=
8g
u
2
So,
S
′
S
=
1
3
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