as this is a projectile you also know the accelration in the vertical direction to be gravity. 9.81 m/s^2.
what you want to do is separate this question into vertical and horizontal. so if you konw the inital velocity, u, at an angle of 50 then you can work out the vertical intial velocity using basic trigonometry.
vertical: you now have vertical inital velocity,u, and you know that when it reaches its heighest point its final velocity, v, will be 0. now you have u, v, a and so you can find the time in the air using v = u + at. once you have found t, double it so you have the total time in the air until it hits the floor and you can use this time for you horizontal calculations.
horizontal: you now have time, t, and you can find the inital velocity again using tigonometry and now just you the equation speed = distance/time to find your projectile distance.
sorry its a bit of a long explanation. i hope it helps
i also think there's a range equation that looks a lil somehting like this
R = [v^2sin2(theta)]/g
but I am not 100% sure about that