IUV

Verified answer

Х=v0*sin(alpha)*t-g*t^2/2
y=v0*cos(alpha)*t
vx=v0*sin(alpha)-g*t
vy=v0*cos(alpha)

скалярное произведение вектора скорости и радиус-вектора в искомой точке равно нулю
r*v=x*vx+y*vy=0
(v0*sin(alpha)*t-g*t^2/2)*(v0*sin(alpha)-g*t)+v0*cos(alpha)*t*v0*cos(alpha)=0
(v0*sin(alpha)-g*t/2)*(v0*sin(alpha)-g*t) + v0^2*cos^2(alpha)=0
(v0*sin(alpha)-g*t/2)*v0*sin(alpha)-(v0*sin(alpha)-g*t/2)*g*t + v0^2*cos^2(alpha)=0
 v0^2=g*t/2*v0*sin(alpha) + (v0*sin(alpha)-g*t/2)*g*t
 v0^2+g^2t^2/2=3*g*t/2*v0*sin(alpha)
g^2t^2/2-3*g*t/2*v0*sin(alpha)+ v0^2=0
t^2-3*t*v0/g*sin(alpha)+ 2(v0/g)^2=0
d=(v0/g)^2*(9*sin^2(alpha)-8)
t1=v0/g*(3*sin(alpha)-корень(9*sin^2(alpha)-8))/2
t2=v0/g*(3*sin(alpha)+корень(9*sin^2(alpha)-8))/2