The motion of the ball is a composition of two motions:
- on the x (horizontal) axis, it is a uniform motion with initial velocity [tex]v_x = 1.0 m/s[/tex]
- on the y (vertical) axis, it is a uniformly accelerated motion with acceleration [tex]g= 9.81 m/s^2 [/tex]
(a) to solve this part, we just analyze the motion on the vertical axis. The law of motion here is
[tex]y(t) = h - \frac{1}{2} gt^2[/tex]
By requiring y(t)=0, we find the time t at which the ball reaches the floor:
[tex]h- \frac{1}{2}gt^2=0 [/tex]
[tex]t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2\cdot 2.0 m}{9.81 m/s^2} }=0.64 s [/tex]
(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):
[tex]x(t) = v_x t = (1.0 m/s)(0.64 s)=0.64 m[/tex]
