A stone thrown at an angle in such a way that the horizontal component of its initial velocity is 20 m/s and the vertical component of its initial velocity is 40 m/s. How far does the stone travel before landing? How high does the stone travel before landing?

When I calculated the time I got 4.08 s
From that I got that the distance traveled in the x direction was 81.63
For the height I got 163.27 Did I do this correctly?

When i calculated the time i got 4.08 s

Not quite - that's the time for the stone to reach its maximum height. It will take another 4.08 second to fall back to the ground. This comes from this formula:


d = d_0 + v_0 t + \frac 1 2 a t^2


where

d = 0\\
d_0 = 0\\
v_0 = 40 \frac m s\\
a = -9.8 \frac m {s^2}


and solve for t.


From that i got that the distance traveled in the x direction was 81.63

No, you need to fix the value for total time.


For the height i got 163.27 Did i do this correctly?

No - not sure how you got that result. The max height can be found in a couple of ways. You can use


d = d_0 + v_0 t + \frac 1 2 a t^2


and set d_0 = 0, v_0 = 40 m/s, and t = 4.08 s. Or you can use


v_2^2 - v_1 ^2 = 2ad

where v_2 = 0 ,\ v_1 = 40 m/s, and solve for d.

Ok so I redid the problem and I got t equal to 9.79 seconds is that correct?

No, could you show how you got that time please?