Consider the case where a projectile is launched horizontally at a speed of 5.0 m/s from a height of 4.0 m

a. Determine the time it takes for the projectile to strike the ground.
T^2 = 2d / a t^2 = 2(-4.0m) / -9.80m/s2 t = 0.9s
b. What is the horizontal component of the projectile’s velocity when it strikes the ground?
Vx = vx1 vx = 5.0m/s
c. What is the vertical component of the projectile’s velocity when it strikes the ground?
V22 = v12 + 2ad v2 = (0m/s2) + 2(-9.80m/s2)(-4.0m) v2 = 8.85m/s
d. Determine the magnitude and direction (down from the horizontal) of the projectile when it strikes the ground.
e. What is the range of the ball?

i got this far but have no idea how to do the last two questions

I find your variables confounding and inserting the units into the equations makes reading the equations even harder. You really should clean up your equations; so people can follow them better and easier.

d. The magnitude of the impact velocity is V = sqrt(Ux^2 + Vy^2); where Ux = vx and Vy = 8.85 mps if I read your factors correctly. Then the impact angle is theta = arctan(Vy/Ux) CAUTION: I'm guessing your V22 should be minus 8.85 mps because that's pointing downward. Check your work.

e. R = Ux T; where Ux = vx = 5 mps and T = .9 sec. You should have known the X speed is constant; so all you needed to do was multiply it by the flight time T to get the range.