Trajectory question? so confused

**The Black Pearl has once again sailed into Port Royal and is firing its guns at the fort. The cannons atop the fort wall have been disabled, but there remains a single cannon at sea level which must defend the town. If the gunner estimates that the ship is 480 meters horizontally from the guns and he knows that the cannon has a muzzle velocity of 87 m/s, at what angle (or angles) should the cannon be aimed so as to hit the Black Pearl? Ignore air resistance, assume **https://services.exchange.deakin.edu...s/clear1x1.gif** and that the cannon and Black Pearl are at the same height above sea level. **

**Note: Enter your answers as a single entry, a comma separated list or, if the cannonball cannot hit the ship, enter 'none'. **

**The cannon can be fired at ???????? degrees to hit the ship. **

so i have got to the point.

a (t)= <0 , -9.8>

v(t)= < 87cos(a) , -9.8t + 87sin(a) >

p(t)= <$\displaystyle 87cos(a)*t$ , $\displaystyle -4.9t^2 + 87sin(a)t$>

and we need the point (480,0)

so $\displaystyle 480 = 87cos(a)*t$

and $\displaystyle 0 = -4.9t^2 + 87sin(a)t$

$\displaystyle t= \frac{480}{87cos(a)}$

therefore$\displaystyle 0 = -4.9[\frac{480}{87cos(a)}]^2 + 87sin(a)[\frac{480}{87cos(a)}]$

$\displaystyle 0 = \frac{-149.16}{cos(a)^2} + 480tan(a)$

but i don't know how to simplify it any further to get an angle?

Re: Trajectory question? so confused

Hey linalg123.

Hint: Use the relationship between sec^2(x) and tan^2(x) where sec^2(x) = 1 + tan^2(x) and get everything in terms of tan(x) and tan^2(x) and solve for x.