Projectile problems

Divide the motion into x and y-components, and deal with them as if you were dealing with one dimensional motion.

The equations:
[tex]\left\{ \begin{array}{l}
x = x_0+v_{x0}t+\frac{1}{2}a_xt^2 \\
y = y_0+v_{y0}t+\frac{1}{2}a_yt^2 \\
\end{array} \right.
[/tex]

 

Dont worry about those position equations. Though they work elegantly and efficiently, there is another, more efficient method. since all you have is the RANGE and the initial velocity, plus the fct that the initial and final height are the SAME, the range equation is your new best friend. By eliminating the time parameter and substitution, you get the more useful eqution: R=V0^2*sin(2theita)/g. the beauty here is that g is a given, along with Vo and R. just solve for theita and rember that there will be two angles, one slightly above and below 45 degrees.

 

True, that is the equation we get for this problem from the position equations. One should not in any mathematical subject learn equations by heart. Instead, one should learn how to derive them.

The position equations give far more information on the projectile's movement than the "range equation".