Whew.
The equation of a trajectory is
y = x tanθ - gx^2/(2*Vo*cos^2(θ))
y=0 ==> x=0
y=10 when x=40 ==>
10 = 40tanθ - 9.8x^2/(2*900*cos^2 θ)
10 = 40tanθ - 8.71sec^2 θ
10 = 40tanθ - 8.71 - 8.71tan^2 θ
8.71tan^2 θ - 40tanθ + 18.71 = 0
solving that for tanθ we get
tanθ = 0.5286 or 4.0638
Letting
a = arctan(.5286)
b = arctan(4.0638)
evaluating tan(a+b) = -4
a particle is projected from a point 0 with an initial speed of 30m/s to pass through a point which is 40m from 0 horizontally and 10m above 0. show that there are two angles of projection for which this is possible. if these angles are alpha and beta show that tan(alpha + beta) = -4
1 answer
1 answer