Question

A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point as seen from the point of projection is ​

Answer 1

$\Huge\underline{\underline{\mathfrak \green{Answer}}}$

Elevation angle of the projectile at its highest point as seen from the point as seen from the point of projection is

${\bf \red{\alpha =tan^{-1}\left(\dfrac{1}{2}\right)}}$

$\Huge\underline{\underline{\mathfrak \green{Solution}}}$

$\large\underline{\underline{\sf Given:}}$

• Angle of projection = 45°

$\large\underline{\underline{\sf To\:Find:}}$

• Elevation angle of the projectile at its highest point as seen from the point as seen from the point of projection is

$\large\underline{\underline{\sf Formula\:Used-}}$

$\large{\boxed{\boxed{\sf \red{Height\:of\: projection (H)=\dfrac{u^2sin^2\theta}{2g}}}}}$

$\large{\boxed{\boxed{\sf \red{ Range\:of\: projection (R)=\dfrac{u^2sin2\theta}{g} }}}}$

❏ Height of Projection

$\Large\implies{\sf \purple{H=\dfrac{u^2sin^2\theta}{2g}}}$

$\implies{\sf \dfrac{u^2sin^245°}{2g}}$

$\implies{\sf \blue{H=\dfrac{u^2}{4g}} }$

❏ Range of Projectile

$\Large\implies{\sf \purple{R=\dfrac{u^2sin2\theta}{g}}}$

$\implies{\sf R=\dfrac{u^2sin90°}{g}}$

$\implies{\sf R=\dfrac{u^2}{g}}$

$\implies{\sf \blue{\dfrac{R}{2}=\dfrac{u^2}{2g}}}$

⛬ ${\sf tan\:\alpha=\dfrac{H}{R/2} }$

$\implies{\sf \dfrac{u^2/4g}{u^2/2g}}$

$\implies{\sf tan\:\alpha=\dfrac{1}{2}}$

$\large\implies{\bf \red{\alpha =tan^{-1}\left(\dfrac{1}{2}\right)}}$

Answer 2

Explanation:

Elevation angle of the projectile at its highest point as seen from the point as seen from the point of projection is

{\bf \red{\alpha =tan^{-1}\left(\dfrac{1}{2}\right)}}α=tan

−1

(

2

1

)

\Huge\underline{\underline{\mathfrak \green{Solution}}}

Solution

\large\underline{\underline{\sf Given:}}

Given:

Angle of projection = 45°

\large\underline{\underline{\sf To\:Find:}}

ToFind:

Elevation angle of the projectile at its highest point as seen from the point as seen from the point of projection is

\large\underline{\underline{\sf Formula\:Used-}}

FormulaUsed−

\large{\boxed{\boxed{\sf \red{Height\:of\: projection (H)=\dfrac{u^2sin^2\theta}{2g}}}}}

Heightofprojection(H)=

2g

u

2

sin

2

θ

\large{\boxed{\boxed{\sf \red{ Range\:of\: projection (R)=\dfrac{u^2sin2\theta}{g} }}}}

Rangeofprojection(R)=

g

u

2

sin2θ

❏ Height of Projection

\Large\implies{\sf \purple{H=\dfrac{u^2sin^2\theta}{2g}}}⟹H=

2g

u

2

sin

2

θ

\implies{\sf \dfrac{u^2sin^245°}{2g}}⟹

2g

u

2

sin

2

45°

\implies{\sf \blue{H=\dfrac{u^2}{4g}} }⟹H=

4g

u

2

❏ Range of Projectile

\Large\implies{\sf \purple{R=\dfrac{u^2sin2\theta}{g}}}⟹R=

g

u

2

sin2θ

\implies{\sf R=\dfrac{u^2sin90°}{g}}⟹R=

g

u

2

sin90°

\implies{\sf R=\dfrac{u^2}{g}}⟹R=

g

u

2

\implies{\sf \blue{\dfrac{R}{2}=\dfrac{u^2}{2g}}}⟹

2

R

=

2g

u

2

⛬ {\sf tan\:\alpha=\dfrac{H}{R/2} }tanα=

R/2

H

\implies{\sf \dfrac{u^2/4g}{u^2/2g}}⟹

u

2

/2g

u

2

/4g

\implies{\sf tan\:\alpha=\dfrac{1}{2}}⟹tanα=

2

1

\large\implies{\bf \red{\alpha =tan^{-1}\left(\dfrac{1}{2}\right)}}⟹α=tan

−1

(

2

1

)

hope it's helpful to you ☺️