Derive an expression for maximum height attained by a projectile
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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
Hope my answer helps you. Please mark me as Brainliest.
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
Hope my answer helps you. Please mark me as Brainliest.
skml73:
You're welcome.
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