Projectile Motion Help

A projectile is launched from ground level at an angle 45 degrees above the horizontal, and lands on a rooftop 60 meters above the ground. The projectile traveled for 3.0 seconds before it landed. With what speed was the projectile launched?

We solve y(t) = h + Uy t - 1/2 g t^2, which defines the height of the projectile at any time t. So we let t = T = 3 sec, the total flight time, h = 0 at ground level launch, and y(T) = 60 m = Uy T - 4.9 T^2 = Uy 3 - 4.9*3^2.

Then Uy = (60 + 4.9*9)/3 = 34.7 mps and U = Uy/sin(45) = 34.7/sin(45) = 49.1 mps ANS.

using y = uyt - 1/2 g t^2 , u get :

60 = Vsin45 x 3 - 1/2 x 9.8 x 9

=> 60 = 2.12 V - 44.1

=> V = 49.1 m/s