A missle is fired with a launch velocity of 4600 m/s at a target 1932 km away, at what angle must it be fired to hit the target? How long the missle is in the air before hitting the target?

Answer 1

the range is R = v^2/g sin2θ
so, you need

4600^2/9.8 sin2θ = 1932000
sin2θ = 0.8947
2θ = 63.48°

now use the fact that

v = 4600sinθ - 9.8t
to get the flight time to max height (when v=0). Double that for the whole trip

Answer 2

a. Range = Vo^2*sin(2A)/g = 1.932*10^6 m.
4600^2*sin(2A)./9.8 = 1.932*10^6,
2.16*10^6*sin(2A) = 1.932*10^6,
sin(2A) = 0.89444.
A = 31.7o.

b. Xo = Vo*Cos A = 1932*Cos 31.7 = 1643.8 m/s. = Hor. component of initial velocity.

Range = Xo*t = 1.932*10^6.
1643.8 * t = 1.932*10^6,
t = 1175 s. = 19.6 min.