Proving Missile Collision Conditions for Target Hit

[danago]

A missile is fired at a target from the origin O, with the velocity vector, t seconds after it was fired, given by [itex]\overrightarrow v (t) = [u\cos \theta ]\overrightarrow i + [u\sin \theta - gt]\overrightarrow j[/itex], where u, theta and g are constants. The target is moving with velocity [itex]v\overrightarrow i[/itex] and at the instant the missile is fired, the target is at position [itex]h\overrightarrow j[/itex].

Prove that for the missile to hit the target [itex]u^2 \ge v^2 + 2gh[/itex]


Alright, from the information given, I've come up with the following set of displacement equations:

[tex]
\begin{array}{l}
\overrightarrow r _{missile} (t) = \left( {\begin{array}{*{20}c}
{ut\cos \theta } \\
{ut\sin \theta - 0.5gt^2 } \\
\end{array}} \right) \\
\overrightarrow r _{t\arg et} (t) = \left( {\begin{array}{*{20}c}
{vt} \\
h \\
\end{array}} \right) \\
\end{array}
[/tex]

For the missile to hit the target, both components of the motion must be equal for the same value of t; that is:

[tex]
\begin{array}{l}
ut\cos \theta = vt \\
ut\sin \theta - 0.5gt^2 = h \\
\end{array}
[/tex]

Now, the first equation is only true for t=0, unless [itex]u\cos \theta = v[/itex], which i interpreted as a requirement for the collision to occur. From the second equation, the time when the vertical components of displacement are equal is give by:

[tex]
t = \frac{{u\sin \theta \mp \sqrt {u^2 \sin ^2 \theta - 2gh} }}{g}
[/tex]

Now its here where I am not really sure what to do. A hint would be greatly appreciated

Thanks,
Dan.

 

[danago]

Ahh, the second i posted this i realized what to do. Since [itex]u\cos \theta = v[/itex], it can be shown that [itex]\sin \theta = \frac{{\sqrt {u^2 - v^2 } }}{u}[/itex], and then i just sub that into the quadratic discriminant and then set it to be greater than or equal to zero. Sound right?

Thanks anyway :P