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?
Calculate range and
2 Answers
The answers are
Explanation:
The trajectory of a projectile in the
Here,
The initial velocity is
The angle is
The acceleration due to gravity is
The points are
Therefore,
Multiply equation
Then,
But,
Therefore,
The range is when
Therefore,
Then,
The range is
-
# " 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
#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
The range is the second root of
#"ie " qquad qquad tan alpha = (dy/dx )_(x = 0) qquad equiv (doty/dotx )\_(t = 0) qquad = y'(0)#