拋體運動 找角度

Let the initial velocity of the projectile be u, making an angle of b with the horizontal.

Horizontal velocity component, Vx = u.cos(b)
At horizontal distance x from the point of projection, we have
x = [u.cos(b)].t where t is the time of travel
i.e. t = x/[u.cos(b)] ------------ (1)

Initial vertical velocity component = u.sin(b)
Apply equation of motion: v = u +at
hence, Vy = u.sin(b) - gt
where Vy is the vertical velocity at time t after projection
Using equation (1), Vy = u.sin(b) - gx/[u.cos(b)]

Let c be the angle of the projectile at distance x
tan(c) = Vy/Vx
i.e. tan(c) = {u.sin(b) - gx/[u.cos(b)]}/(u.cos(b)
tan(c) = tan(b) - gx/(u.cos(b))^2

The value of angle c can be found if values of u (initial projection speed), b (angle of projection), and x (distance from projection point).

可以用tan去計
先找出水平速率(Vx)及垂直速率(Vy)
(Vy)/(Vx)=tan角度