A bullet fired from a rifle attains maximum height of 25 m and crosses a range of 200m. Find the angle of projection.
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A bullet fired from a rifle attains maximum height of 25 m and crosses a range of 200m. Find the angle of projection.
Plz friends give me correct answer
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Given :-
[tex]H = 25 m[/tex]
[tex]R = 200 m[/tex]
[tex]g=9.8 \frac{m}{s^2}[/tex]
Required to find ;-
θ = ?
Procedure ;-
Let us recall the maximum height achieved by projectile as follows :
[tex]H=\frac{u^2sin^2 \theta}{2g}[/tex]
Let us plug in all that we know.
[tex]25=\frac{u^2sin^2 \theta}{2*9.8}[/tex]
[tex]490=u^2sin^2 \theta[/tex] --------- 1
Now let us recall the range achieved by projectile as follows :
[tex]R=\frac{u^2sin2\theta}{g}[/tex]
Let us now plug in all that we know.
[tex]200=\frac{u^2sin2\theta}{9.8}[/tex]
[tex]1960=u^2sin2\theta[/tex] ---------- 2
Now let us divide 2 by 1 .
[tex]\frac{1960}{490} = \frac{sin^2\theta}{sin2\theta}[/tex]
[tex]4 = \frac{sin\theta}{2cos\theta}[/tex]
[tex]8=tan\theta[/tex]
[tex]\theta=tan^{-1}(x)[/tex]
∴Angle of projection is [tex]tan^{-1}(x)[/tex] .
Verified answer
Given :-
H = 25 m
R = 200 m
Required to find ;-
θ = ?
Procedure ;-
Let us recall the maximum height achieved by projectile as follows :
Let us plug in all that we know.
25={u^2sin^2 \theta}{2*9.8}25=2∗9.8u2sin2θ
490=u^2sin^2 \theta490=u2sin2θ --------- 1
Now let us recall the range achieved by projectile as follows :
R={u^2sin2\theta}{g}R=gu2sin2θ
Let us now plug in all that we know.
200={u^2sin2\theta}{9.8}200=9.8u2sin2θ
1960=u^2sin2\theta1960=u2sin2θ ---------- 2
Now let us divide 2 by {1960}{490} = \frac{sin^2\theta}{sin2\theta}4901960=sin2θsin2θ
4 = \frac{sin\theta}{2cos\theta}4=2cosθsinθ
8=tan\theta8=tanθ
\theta=tan^{-1}(x)θ=tan−1(x)
∴Angle of projection is tan^{-1}(x)tan−1(x)
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