Answer:
1) Distance traveled equals 23.1 meters.
2) Final velocity equals 21.658 m/s.
Explanation:
The problem can be solved using second equation of kinematics as
[tex]s=ut+\frac{1}{2}at^{2}[/tex]
where
s is the distance covered
u is the initial speed of the ball
a is the acceleration the ball is under
t is time of travel
Applying the given values in the above equation we get
[tex]s=4.0\times 1.8+\frac{1}{2}\times 9.81\times 1.8^{2}\\\\s=23.1meters[/tex]
Part 2)
The velocity after 't' time can be obtained using first equation of kinematics.
[tex]v=u+at[/tex]
Applying the given values we get
[tex]v=4+9.81\times 1.80\\\\\therefore v=21.658m/s[/tex]
