Answer:
t=0.816 s
y=3.26 m
Explanation:
Vertical Launch
If an object is thrown vertically in free air (no friction considered), it starts to move upwards at its maximum speed and the acceleration of gravity acts to brake it. At a given time, the object stops in mid-air and starts to fall back to the launching point.
The height of the ball at any time t is given by
[tex]\displaystyle y=vo.t-\frac{gt^2}{2}[/tex]
Where vo is the launching speed, g is the acceleration of gravity and t is the time
The speed of the ball is
[tex]v_f=v_o-g.t[/tex]
a) The ball will keep going upwards until it runs out of speed. It happens when
[tex]v_f=v_o-g.t=0[/tex]
Solving for t
[tex]\displaystyle t=\frac{v_o}{g}[/tex]
[tex]\displaystyle t=\frac{8}{9.8}[/tex]
[tex]t=0.816\ s[/tex]
b) The maximum height is achieved at the time found above:
[tex]\displaystyle y=(8).(0.816)-\frac{(9.8)(0.816)^2}{2}[/tex]
[tex]\boxed{y=3.26\ m}[/tex]
