Q :- A body projected vertically upwards crosses points A and B separated by 28 m with velocity one-third and one-fourth of the initial velocity , respectively. What is the maximum height reached by it above the ground ??
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Answers
Answered by
1
Step-by-step explanation:
Velocity of the body at point A =
3
u
Velocity of the body at point B =
4
u
Apply v
2
−u
2
= 2aS for points A and B:
(
4
u
)
2
−(
3
u
)
2
=2(−g)×28
On solving we get: u
2
= 1152g
Now apply direct formula for maximum height
H
max
=
2g
u
2
H =
2g
1152g
= 576 m
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Answered by
2
Step-by-step explanation:
Let initial velocity of the body = u
Velocity of the body at point A =
3
u
Velocity of the body at point B =
4
u
Apply v
2
−u
2
= 2aS for points A and B:
(
4
u
)
2
−(
3
u
)
2
=2(−g)×28
On solving we get: u
2
= 1152g
Now apply direct formula for maximum height
H
max
=
2g
u
2
H =
2g
1152g
= 576 m
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