Question

13. If u=aî+bj^+ck^ with i^, j^,k^are in east,
north and vertical directions, the maximum
height of the projectile is
1)a^2/2g 2)b^2/2g 3)c^2/2g 4)b^2c^2/2g​

Answer 1

Answer:

Option (C)

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Answer 2

We have to find Height projectile for Vertical direction

We are given

Intial velocity u =  ai^+ bj^ +ck^

Accleration is due to free fall and hence "g" will be considered and is in the Z axis so

acceleration = 0i + 0j - gk

                    = -gk

Now for height

Intial velocity (u) = 0

Final velocity (v) = c

So,

$v^{2} - u^{2} = 2gs$

$c^{2} - 0 = 2gs$

$s = \frac{c^{2} }{2g}$

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