The motion of the ball is a composition of two motions:
- on the x (horizontal) axis, it is a uniform motion with initial velocity

- on the y (vertical) axis, it is a uniformly accelerated motion with acceleration
(a) to solve this part, we just analyze the motion on the vertical axis. The law of motion here is

By requiring y(t)=0, we find the time t at which the ball reaches the floor:


(b) for this part, we can analyze only the motion on the horizontal axis. To find how far the ball will land, we must calculate the distance covered on the x-axis, x(t), when the ball reaches the ground (so, after a time t=0.64 s):
Is there supposed to be a picture? Next time try putting a picture
No, because sometimes you have to stop at stop signs and stop lights.
The correct response that will be used to describe this particular element would be the third option, since all of the other options are incorrect and apply to different elements in their groups. The element is a metal and will react with a non metal.