Find angle of release

[Superman2]

A cannon that shoots projectiles at 100m/s must hit a target 400m away and 30m above the ground. What angle must the cannon point at in order to hit the target?
I have two functions:
x = V*cos(theta)*t, where V = initial velocity, and
y=V*sin(theta)*t - (g*t^2)/2 where g = gravity's acceleration
but i can't solve for theta.
can somebody help me?

[HallsofIvy]

Why can't you solve for theta?

Since you are trying to hit a target 400 m away with initial speed 100 m/s, the horizontal equation, x = V*cos(theta)*t, becomes
400= 100 cos(theta)t or t= 40/cos(theta).

Since the height of the target is 30 m above the ground (which, I assume, is the height of the cannon), the vertical equation,
y=V*sin(theta)*t - (g*t^2)/2, becomes 30= 100 sin(theta) t- (g t²)/2. Now using t= 40/cos(theta), we can write that as 30= 4000 (sin(theta)/cos(theta))- 800g/cos²(theta).

Multiplying both sides by cos²(theta),
30 cos²(theta)= 4000 sin(theta)cos(theta)- 800g

That's a little complicated but can be solved.

[Superman2]

cool i solved it, it was pretty damn long

[slain]

great :) this kinda answeres the first part of my question earlier...

[slain]

:| just one question... the variable t in that equation is time... correct?

[Superman2]

yes t is time

look here
http://www.intel.com/education/unitplans/physics/lessonplans/physics_motion.htm

it shows it in the last solution in the page.

[slain]

ok... well this kind of explains a few things about designing the trajectory of a bullet fired from a weapon at a given angle. let me see if i have this right.
If a bullet is fired at a specific angle and muzzle velocity then it would be at point x and y which can be derived from the equations at time t right?

like if V = 1200 m/s and angle of fire is 60* then the bullet would be at point (12000, 18824) right?

[Superman2]

The solutions in the beginnins solve different stuff including t i think.

in the last solution, if the bullet initial velocity is 1000m/s and the target is at point (3000, 4000) then, the equation solves the firing angle.

Its a little hard to explain what it does but here's a tip. read it over 3-4 times and ull prob get it.

so basically it solves for theta in the equations x = (Vx)t, and
y = (Vy)t + (1/2)at^2

so first it solves algebraicaly for t in the first equation which becomes t = x / (Vx), which is 3000m/Vx or 3000m/(1000cos(theta))

then it substitutes it for t in the second equation in the third line of the solution.
then it reduces it in the fourth line, then multiplies both sides by cos^2(theta)
then it reduces again which results in the fifth line and then
substitutes sin(theta) for sqrt(1 - cos^2(theta))
then after squaring both sides and reducing the equation becomes a second degree one like ax^2 + bx + c = 0 , where x is cos^2(theta)
then it uses the quadric formula
x = ( -b -+ sqrt(b^2 - 4ac) ) / (2a) to find cos^2(theta) then from there its easy it find the square root of the answer and then finds the cos-1 of the answer to find theta;
lol thats long, a friend told me theres a much easier way in advanced physics school books.