Answer:
The ball never passes 2m high, Hmax=1.27m
Step-by-step explanation:
we assume the ball doesn't bounce when it hits the ground.
We calculate the maximum height, Vf = 0.
![v_(o)^(2)=2gH_(max)\\H_(max)=v_(o)^(2)/(2g)=5^(2)/(2*9.81)=1.27m](https://img.qammunity.org/2020/formulas/physics/college/5m7bbd0zg8uy0eswrfky4zaafluunhzrpt.png)
So, the ball never passes 2m high.
Kinematics equations:
![x(t)=v_(o)t-1/2*g*t^(2)\\v(t)=v_(o)-gt](https://img.qammunity.org/2020/formulas/physics/college/2ahqv99bfbigpxxv6zyrr3rml91avns5ma.png)
Find annexed the graphics of x(t) and v(t)