# Projectile problems

1. Sep 26, 2005

### garboui

i have been give an asignment to do in physics. well everything is fine and dandyuntill the last question.

what the question wants to know is that for a projectile launched at a certain velocity(82m/s) hast to reach a set distance(560m), find the two angles that will allow the target distance to be reached. takeoff and landing are at the same height.

is there some formula where i could plug all of the numbers in and get the answer or is there something that i am missing here. i know its easy but for some reason i just cant figure it out for the life of me.

any help will be greatly appreciated as the assignment is due tommrow morning.

thanks
garboui

2. Sep 26, 2005

### Päällikkö

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

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

3. Sep 26, 2005

### DaMastaofFisix

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.

4. Sep 27, 2005

### Päällikkö

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".