Arun
Last Activity: 6 Years ago
The horizontal speed is the same at any time in the trajectory.
u = total speed at 1/2 max height
v = horizontal speed = u*sqrt(2/5)
w = vertical speed at 1/2 max height. Use c^2 = a^2 + b^2 to find an expression for w.
u^2 = u^2*(2/5) + w^2
w^2 = u^2*(1 - 2/5)
w = u*sqrt(3/5)
We now need to find the initial vertical speed.
H = maximum height
H/2 = (1/2)*g*t^2
H = (1/2)*g*T^2
T/t = sqrt(2)
w = g*t
W = g*T .... W = w*T/t = w*sqrt(2) = u*sqrt(6/5)
tan (Theta) = W/v = sqrt(6/5)/sqrt(2/5) = sqrt(3)
So theta is 60 degrees
