# Projectile Motion

1. Oct 15, 2007

### danago

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

Prove that for the missile to hit the target $u^2 \ge v^2 + 2gh$

Alright, from the information given, ive come up with the following set of displacement equations:

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

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

$$\begin{array}{l} ut\cos \theta = vt \\ ut\sin \theta - 0.5gt^2 = h \\ \end{array}$$

Now, the first equation is only true for t=0, unless $u\cos \theta = v$, 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:

$$t = \frac{{u\sin \theta \mp \sqrt {u^2 \sin ^2 \theta - 2gh} }}{g}$$

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

Thanks,
Dan.

2. Oct 15, 2007

### danago

Ahh, the second i posted this i realised what to do. Since $u\cos \theta = v$, it can be shown that $\sin \theta = \frac{{\sqrt {u^2 - v^2 } }}{u}$, 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