Question

derive the equation for Maximum height attained by the projectile in case

of oblique projection.​

Answer 1

MAXIMUM HEIGHT

• At maximum height position (B), $\sf{v_y=0}$
• Acceleration = -g
• Initial velocity = $\sf{u_y}$ = u sinθ

If +H is the maximum height, then :

$\sf{(v_y)^2=(u_y)^2+2(-g)H}$

$\sf{\longrightarrow (0)^2=(u\ sin\theta)^2-2gH}$

$\sf{\longrightarrow 0=u^2sin^2\theta-2gH}$

$\sf{\longrightarrow 2gH = u^2sin^2\theta}$

$\boxed{\sf{\longrightarrow H=\dfrac{u^2sin^2\theta}{2g} }}$

The above boxed equation is the equation for maximum height attained by the projectile in case  of oblique projection.​

Answer 2

To do:

• Derivation of the equation for maximum height attained by the projectile in case of oblique projection.

Solution:

In an oblique projection,

Horizontal component of velocity, $\sf{u_x}$ = u cos θ

Vertical component of velocity, $\sf{u_y}$ = u sin θ

Acceleration due to gravity, a = -g

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here,

Let, the maximum height attained by projectile be H

so, vertical velocity of Projectile at maximum height; (v) will be zero

taking acceleration due to gravity, a = -g ; and

Initial velocity of projectile = vertical component of velocity = u sin θ

[ Where θ is angle of projection ]

then,

By third equation of motion

$\implies\sf{v^2-u^2=2as}$

[ Where v is final velocity, u is initial velocity, a is acceleration, s is distance covered]

$\implies\sf{(0)^2-(u\;sin\;\theta)^2=2(-g)(H)}$

$\implies\sf{0-u^2\;sin^2\theta=-2\;g\;H}$

$\implies\sf{2\;g\;H=u^2\;sin^2\;\theta}$

$\implies\boxed{\sf{H=\dfrac{u^2\;sin^2\theta}{2\;g}}}$

Refer to the attachment  for figure.