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

A) range #R = (a^2 + ab + b^2)/(a+b)#
B) angle of projection #tanalpha = [(a+b+c)/a]#

2 Answers
Cesareo R.
Feb 17, 2018

See below.

Explanation:

The generic path is

#y(x) = c_1 x+c_2 x^2# so

Now solving for #c_1,c_2#

#b = y(a) = a c_1 + a^2 c_2 #
#a = y(b) = b c_1 + b^2 c_2#

we get at

#c_1 = -(a^2+ab+b^2)/(ab)#
#c_2 = -(a+b)/(ab)#

and then

#y(x) = (1+a/b+b/a)x-x^2/(a+b)#

and the range is obtained for #y(x)=0# giving

#R = (a^2 + a b + b^2)/(a + b)#

and the projection angle is given by

#y'(0) = tan alpha = 1 + a/b + b/a#

Narad T.
Feb 17, 2018

See the explanation below

Explanation:

Any point #(x,y)# on the trajectory of the projectile is given by the equation

#y=-(gx^2)/(2v_0^2cos^2alpha)+x tanalpha#

As the points #(a,b)# and #(b,a)# belong to the trajectory

Therefore,

#b=-(ga^2)/(2v_0^2cos^2alpha)+a tanalpha#...........#(1)#

#a=-(gb^2)/(2v_0^2cos^2alpha)+b tanalpha#...........#(2)#

Solving for #tanalpha# in equations #(1)# and #(2)#

#atanalpha-b=(ga^2)/(2v_0^2cos^2alpha)#...........#(3)#

#btanalpha-a=(gb^2)/(2v_0^2cos^2alpha)#..............#(4)#

Therefore,

#(atanalpha-b)/a^2=(btanalpha-a)/b^2#

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

#tanalpha(ab(b-a))=b^3-a^3=(b-a)(a^2+ab+b^2)#

#tanalpha=(a^2+ab+b^2)/(ab)#

Substituting this value in #(3)#

#(2v_0^2cos^2alpha)/g=(a^2)/(a*(a^2+ab+b^2)/(ab)-b)#

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

So,

#(v_0^2/g)=(a^2b)/(2(a^2+ab)*cos^2alpha)#

The range is

#R=(v_0^2)/g*2sinalpha*cosalpha#

#=(a^2b)/(2(a^2+ab)*cos^2alpha)*2sinalpha*cosalpha#

#=(ab)/(a+b)*tanalpha#

#=(ab)/(a+b)*(a^2+ab+b^2)/(ab)#

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

Hi physicslover, you added a #c# in #tanalpha# which is unnecessary. Everything fits well with rhw value of #tanalpha#.

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