Question

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?

Answers

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
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.

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