Projectile Angle Help: Solving for Launch Angle with Initial Velocity of 20m/s

[Petkovsky]

Homework Statement

http://filip2.mkprovider.com/joomla/images/fizika.png

So, we have a body that is launched at a certain angle with initial velocity of 20m/s.
The body is supposed to hit an object, which is 20m further away (horizontally), at the height of 5m. The question is at what angle should the body be launched.

Homework Equations

The equation for the horizontal coordinate is which is 20 is X=v*cos(alpha)*t ,
while the one for the vertical coordinate which is 5 is Y=v*sin(alpha)*t - (1\2)*g*t^2.

The Attempt at a Solution

I expressed time(t) from the first equation and got t=X/v*cos(alpha), substituted it in the second equation, and received Y=v*sin(alpha)*X/v*cos(alpha) -5 * (x/v*cos(alpha))^2, with Y being 5 and X 20.

However after simplifying the equation i received 4*tan(alpha) - sec^2(alpha) - 1 = 0, from where i could not express the angle alpha, therefore i could find its value.

If you have any suggestion of how to solve the problem please assist me.

 

[Doc Al]

Hint: Use a trig identity to rewrite \sec^2\alpha in terms of \tan^2\alpha.

 

[Petkovsky]

Of course...how could I have missed that...thanks a lot, i solved it :).