A ball is thrown at an angle of 53 degrees above the horizontal off the roof (check my answers)?

Question

A ball is thrown at an angle of 53 degrees above the horizontal off the roof of a building with an initial speed of 15 m/s. Find the position and velocity of the ball at 1.0s, 2.0s and 4.0s after it was thrown (ignore air resistance) (x- components are horizontal y components are vertical)

What is the x, y, Vx, Vy and V for the ball at 1.0s, 2.0s and 4.0s

I used equation V=at+Vo to solve for V and X-Xo=.5(V+Vo)t to solve for x.

I got 9m/s for Vox, 12m/s for Voy, ax = -9.8m/s^2 and ax= 0m/s^2

Time = 1.0s, x = 9m, y = 7.1m, Vx= 9m/s, Vy =2.2m/s and V = 15 m/s

Time = 2.0s, x = 18m, y = 4.4m, Vx= 9m/s, Vy =-7.6m/s and V = 15 m/s

Time = 4.0s, x = 9m, y = -15.2m, Vx= 9m/s, Vy =-27.2m/s and V = 15 m/s

Answer 1

I'm afraid that you got off on the wrong track and this is rather obvious because you haven't involved the angle. You need to start by considering sideways motion and upwards motion separately. This is called resolving the velocity into its components.
Vx = Vo*cos53 unchanging because we assume that the air doesn't slow it down. Vy = (Vosin53) + gt where g = acceleration due to gravity, usually taken as -9.8m/s^2. These formulas will give you Vx and Vy at the times you want. V is not so easy but you can get it from the previous two using trigonometry or pythagoras. For x and y you will need to integrate the velocity functions separately. You can find the constant of integration by knowing where the object started. Try to do it again yourself before copying anyone's answer left here.
(I have noticed an alarming number of wrong answers posted on the mathematics part of this web site.)

Answer 2

The velocity vector has components Vx= 15 cos 53 and Vy 15 sin53
Vx=9.03 m/s and is constant
Vy (initial) =11.98m/s So the equation of the vertical movement is

Vy =11.98-9.8t and Y = -9.8/2 t^2 +11.98t (reference the roof of the buiding)
V(1s) = sqrt(9.03^2 +(11.98-9.8)^2)) = 9.29 m/s
Y(1s) = 7.08 m above the roof and 9.03m distant horizontally
V(2s) = sqrt(9.03^2 +(11.98-19.6)^2)=11.82 m/s(pointing downwards)
Y(2s) = 4.36m above the roof and 18.06m distant horizontally

Now you can do it for 4 s