A particle is projected at a definite angle $α$ $alpha$ to the horizontal passes through the points $(a, b)$ $(a,b)$ and $(b, a)$ $(b,a)$ , referred to horizontal and vertical axes through the point of projection?

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A particle is projected at a definite angle $α$ $alpha$ to the horizontal passes through the points $(a, b)$ $(a,b)$ and $(b, a)$ $(b,a)$ , referred to horizontal and vertical axes through the point of projection?

Calculate range and $tan α$ $tanalpha$ in terms of a and b?

2 Answers

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Answer 1 · 2018-08-09T06:36:21

The answers are $tan α = \frac{b^{2} + a b + a^{2}}{a b}$ $tanalpha=(b^2+ab+a^2)/(ab)$ and the range is $= (a + b)$ $=(a+b)$

Explanation:

The trajectory of a projectile in the $x - y$ $x-y$ plane is given by the equation

$y = x tan θ - (\frac{g}{2 u^{2} {cos}^{2} θ}) x^{2}$ $y=xtantheta-(g/(2u^2cos^2theta))x^2$

Here,

The initial velocity is $= u$ $=u$

The angle is $θ = α$ $theta=alpha$

The acceleration due to gravity is $= g$ $=g$

The points are $= (a, b)$ $=(a,b)$ and $= (b, a)$ $=(b,a)$

Therefore,

$b = a tan α - (\frac{g}{2 u^{2} {cos}^{2} α}) a^{2}$ $b=atanalpha-(g/(2u^2cos^2alpha))a^2$ ............. $(1)$ $(1)$

$a = b tan α - (\frac{g}{2 u^{2} {cos}^{2} α}) b^{2}$ $a=btanalpha-(g/(2u^2cos^2alpha))b^2$ ............. $(2)$ $(2)$

Multiply equation $(1)$ $(1)$ by $b^{2}$ $b^2$ and equation $(2)$ $(2)$ by $a^{2}$ $a^2$

$b^{3} = a b^{2} tan α - (\frac{g}{2 u^{2} {cos}^{2} α}) a^{2} b^{2}$ $b^3=ab^2tanalpha-(g/(2u^2cos^2alpha))a^2b^2$ ............. $(3)$ $(3)$

$a^{3} = a^{2} b tan α - (\frac{g}{2 u^{2} {cos}^{2} α}) b^{2} a^{2}$ $a^3=a^2btanalpha-(g/(2u^2cos^2alpha))b^2a^2$ ............. $(4)$ $(4)$

Then,

$(3) - (4)$ $(3)- (4)$ is

$b^{3} - a^{3} = a b^{2} tan α - a^{2} b tan α$ $b^3-a^3=ab^2tanalpha-a^2btanalpha$

$= a b (b - a) tan α$ $=ab(b-a)tanalpha$

But,

$b^{3} - a^{3} = (b - a) (b^{2} + a b + a^{2})$ $b^3-a^3=(b-a)(b^2+ab+a^2)$

Therefore,

$tanalpha=(b^3-a^3)/(ab(b-a))=(cancel(b-a)(b^2+ab+a^2))/(abcancel(b-a))$

$=(b^2+ab+a^2)/(ab)$

The range is when

$a=0$ or $b=0$

Therefore,

$0=atanalpha-(g/(2u^2cos^2alpha))a^2$ ............. $(3)$

$0=btanalpha-(g/(2u^2cos^2alpha))b^2$ ............. $(4)$

Then,

$(4)- (3)$ is

$(b-a)tanalpha=(b^2-a^2)(g/(2u^2cos^2alpha))$

$u^2=(g(a+b))/(2tanalphacos^2alpha)=(g(a+b))/(sin2alpha)$

The range is

$r=u^2sin(2alpha)/g=(g(a+b))/(sin2alpha)*sin(2alpha)/g=a+b$

Answer 2 · 2018-08-09T18:00:04

$" Range is: " qquad( a^2 + b^2 + ab )/( a+b )$
$qquad tan alpha = (a^2 + b^2 + ab)/(ab)$

Explanation:

Projectile motion is parabolic . This is easy to show, by replacing parameter $t$ , in the equations for horizontal and vertical displacements, by the fixed values, namely $u_x, u_y, alpha$ :

$qquad qquad {(x(t) = u_x t),(y(t) = u_y t - 1/2 g t^2),(tan alpha = u_y / u_x):}$

So for parabola passing through origin:

$y = x ( lambda x + mu) qquad " Roots " to qquad {(x = 0),(x = bb(- mu/lambda) qquad square):}$

It follows from the IV's that:

${(b = a ( lambda a + mu)),(a = b ( lambda b + mu) ):} qquad " or " qquad [(a,1),(b,1)][(lambda),(mu)] = [(b/a),(a/b)]$

Noting the restrictions that make the algebra impossible, or the matrix singular (esp $a ne b$ ), this leaves:

$lambda = - (a+b)/(ab) qquad qquad mu = (a^2 + b^2 + ab)/(ab)$

The range is the second root of $y(x)$ , as stated in $square$ above:

$:. " Range is: " ( a^2 + b^2 + ab )/( a+b )$

$tan alpha$ is the derivative (wrt $x$ ) of the parabola at the Origin:

$"ie " qquad qquad tan alpha = (dy/dx )_(x = 0) qquad equiv (doty/dotx )_(t = 0) qquad = y'(0)$

$y' = 2 lambda x + mu$

$y'(0) = mu = bb( (a^2 + b^2 + ab)/(ab) = tan alpha)$

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A particle is projected at a definite angle $α$ $alpha$ to the horizontal passes through the points $(a, b)$ $(a,b)$ and $(b, a)$ $(b,a)$ , referred to horizontal and vertical axes through the point of projection?

Calculate range and $tan α$ $tanalpha$ in terms of a and b?

2 Answers

Explanation:

Explanation:

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Explanation:

Explanation:

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A particle is projected at a definite angle $α$ $alpha$ to the horizontal passes through the points $(a, b)$ $(a,b)$ and $(b, a)$ $(b,a)$ , referred to horizontal and vertical axes through the point of projection?

Calculate range and $tan α$ $tanalpha$ in terms of a and b?