Question

derive an expression for the maximum height attained by projectile.

Answer 1

let a projectile moves with u velocity which inclined with horizontal ∅ angle .
then,
velocity vector after t time (V)
V = Vx i + Vy j
V = ucos∅ i + (usin∅ -gt) j

at maximum height velocity of projectile have only x -components exist .e.g
Vx =ucos∅
use , Vy² = Uy² + 2ay.Y
for projectile
Vy at maximum height = 0
Uy = usin∅
ay = -g
Y = maximum height

put this ,
0 = (usin∅)² -2gHmax

Hmax = u²sin²∅/2g

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Answer 2

Answer:

$when \: an \: object \: is \: thrown \\ \: upwards \\ \\ final \: velocity = v = 0ms {}^{ - 2} \\ \\ acceleration \: due \: to \: \\ gravity = - g \\ \\ using \: newtons \: third \: \\ equation \: of \: motion \\ v {}^{2} = u {}^{2} + 2gs \\ 0 {}^{2} = u {}^{2} + 2( - g)s \\ 0 = u {}^{2} - 2gs \\ 2gs = u {}^{2} \\ s = \frac{u {}^{2} }{2g} \\ \\ therefore \: maximum \: height \: \\ attained \: by \: an \: object \: can \: be \: \\ found \: by \: the \: formula \\ \\ s = \frac{u {}^{2} }{2g}$

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