Projectile motion question?

Vertical Component of Velocity = V sin θ

Vertical Velocity at highest point = 0

Maximum Height attained be h

Acceleration = -g

[negative sign because direction of acceleration is opposite to direction of velocity]

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Time to reach max height ---->

v = u + at [Ist Equation of motion]

0 = V sin θ - gt

=> t = (V sin θ / g)..............................( i )

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Max height attained ------>

h = ut + 1/2 at² [IInd Equation of Motion]

= (V sin θ)(t) + 1/2 (-g)(t²)

= (V sin θ)(V sin θ / g) - 1/2 (g)(V sin θ / g)² [Use value of t from ( i )]

= [(V sin θ)²] { 1/g - 1/2g } [Take (V sin θ)² as common]

= [(V sin θ)²] { 1/2g }

= (V sin θ)² / 2g or (V²sin²θ) / 2g

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Great answer from Nano Techman.

Using v² = u² + 2as.......where u = Vsinθ, a = -g and s is the vertical distance.

At the greatest height velocity is zero.

=> 0 = (Vsinθ)² - 2gs

=> 2gs = V²sin²θ

i.e. s = V²sin²θ/2g

:)>