Question

Derive the expression for projectile motion time of maximum height

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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