Answer:
The ball never passes 2m high, Hmax=1.27m
Explanation:
we assume the ball doesn't bounce when it hits the ground.
We calculate the maximum height, Vf = 0.
[tex]v_{o}^{2}=2gH_{max}\\H_{max}=v_{o}^{2}/(2g)=5^{2}/(2*9.81)=1.27m[/tex]
So, the ball never passes 2m high.
Kinematics equations:
[tex]x(t)=v_{o}t-1/2*g*t^{2}\\v(t)=v_{o}-gt[/tex]
Find annexed the graphics of x(t) and v(t)
