plz answer the question fast
Answers
The answer is c both a and b
I hope it helps,
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
)
Explanation:
hope it's helpful to you ☺️
