In phyics, section Projectile motion.... and my question

Suppose the projectile is a home-run ball hit with an initial speed v0 = 50m/s at an initial angle θ0 = 53.10. The ball is probably struck a meter or so above ground level, but we neglect this distance and assume it starts at ground level (y0=0). Then
If the ball does not hit the ground, it continues to travel on below its original level. Maybe there is a big hole where the landing point ought to be, or the playing field might be located atop a flat-topped hill that drops off steeply on one side. Then negative values of y, corresponding to times greater than 8.16 s, are possible. Can you compute the position and velocity at a time 10s after the start.
Explain it, pls.

Thanks for your reply!!

Find the x and y components of the initial velocity:

v0(x) = v0 cos (theta) = 50 cos 53.10 = 30 m/s

v0(y) = v0 sin (theta) = 50 sin 53.10 = 40 m/s

Hmm, convenient!

The acceleration is -9.8 m/s/s

The vertical position is given by:

d = ut + 1/2 at^2 = 40*10 + (-9.8)*100 = -90 (that is, 90 m below the starting point)

The horizontal position is simply v = d/t, d = vt = 30* 10 = 300 m

So the position will be found using Pythagoras' theorem to find the displacement from the origin and an inverse tan to find the angle...

The velocity will be the vector sum of the x component of the initial velocity (unchanged) and the y velocity vector given by:

v = u + at = 40 + (-9.8)*10 = -58 m/s

Its direction would be given by that vector sum.