Solving the Displacement of a Catapulted Stone

[rover_dude]

Homework Statement

A stone is catapulted at time t = 0, with an initial velocity of magnitude 20.5 m/s and at an angle of 41.4° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.16 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.70 s.

I tried multiple solutions, but none of them are right, so I do not know what I am doing wrong.

The Attempt at a Solution

Ok so I think this is the closest I got to the answer, but I used the equation V=Vo(t)+(1/2)(a)(t)^2 So for the horizontal component at 1.16s I got x=20.5(1.16)+(1/2)(0)(1.16)^2 and I got x=23.78m. For the vertical component I got y=20.5(1.16)+(1/2)(-9.8)(1.16)^2 and y=17.18656m. For 1.7 seconds: x=20.5(1.7)+(1/2)(0)(1.7)^2 and I got x=34.85m for y: y=20.5(1.7)+(1/2)(-9.8)(1.7)^2 and y=20.689.
(I think these were the closest answers I got, but I tried different methods which were all wrong so help is appreciated)

 

[NascentOxygen]

So for the horizontal component at 1.16s I got x=20.5(1.16)+(1/2)(0)(1.16)^2

The horizontal component of initial velocity is not 20.5

For the vertical component I got y=20.5(1.16)+(1/2)(-9.8)(1.16)^2

The vertical component of initial velocity is not 20.5

 

[rover_dude]

So, If I have the correct initial velocity, this method would be correct?

 

[NascentOxygen]

So, If I have the correct initial velocity, this method would be correct?

Yes.