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

Calculate range and #tanalpha# in terms of a and b?

2 Answers
Aug 9, 2018

Answer:

The answers are #tanalpha=(b^2+ab+a^2)/(ab)# and the range is #=(a+b)#

Explanation:

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

#y=xtantheta-(g/(2u^2cos^2theta))x^2#

Here,

The initial velocity is #=u#

The angle is #theta=alpha#

The acceleration due to gravity is #=g#

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

Therefore,

#b=atanalpha-(g/(2u^2cos^2alpha))a^2#.............#(1)#

#a=btanalpha-(g/(2u^2cos^2alpha))b^2#.............#(2)#

Multiply equation #(1)# by #b^2# and equation #(2)# by #a^2#

#b^3=ab^2tanalpha-(g/(2u^2cos^2alpha))a^2b^2#.............#(3)#

#a^3=a^2btanalpha-(g/(2u^2cos^2alpha))b^2a^2#.............#(4)#

Then,

#(3)- (4)# is

#b^3-a^3=ab^2tanalpha-a^2btanalpha#

#=ab(b-a)tanalpha#

But,

#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#

Aug 9, 2018

Answer:

  • # " 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)#