I assume what you mean by "trajectory" is you need formulas for x- and y-coordinates as functions of time, and same with the angle, correct?
If we ignore air resistance, then given initial velocity V_0, the horizontal velocity is $\displaystyle v_x = V_0 \cos (\theta )$, where $\displaystyle \theta$ is the launch angle. Notice that this is constant. The vertical velocity changes with time as gravity causes the arrow to decelerate, and is $\displaystyle v_y = V_0 \sin (\theta )- gt$ where g is the acceleration dues to gravity: $\displaystyle g = 9.81 \frac m {s^2}$. If the (x,y) coordinate of the launch point is $\displaystyle (X_0,Y_0)$, then the x- and y-positions as a function of time are:
$\displaystyle x(t) = X_0 + V_0 \cos (\theta) t$
$\displaystyle y(t) = Y_0 + V_0 \sin(\theta) t - \frac 1 2 g t^2$
The angle of the arrow to the ground is equal to the arc tangent of the vertical velocity divided by the horizontal velocity (this assumes that the arrow always points in the direction of travel, which is not absolutely correct, but should be close enough for your purposes):
$\displaystyle \alpha = \tan^{-1} ( \frac {v_y}{v_x} ) = \tan^{-1} ( \frac {V_0 \sin (\theta )- gt}{V_0 \cos (\theta)} )$
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Last edited by ChipB; Sep 11th 2017 at 06:34 AM.
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