To solve this problem, we can use the equations of motion for an object under constant acceleration. In this case, the ball is thrown upward with an initial velocity and experiences the acceleration due to gravity.
Let's denote the time as 't' seconds and the initial velocity as 'u' (which is 13 m/s in this case).
We know that the height above the ground is the vertical displacement of the ball. Using the equation for vertical displacement:
s = ut + (1/2)at^2
where:
s = vertical displacement (unknown in this case)
u = initial vertical velocity (13 m/s)
a = acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
t = time in seconds (2 seconds in this case)
Plugging in the values, we have:
s = (13)(2) + (1/2)(-9.8)(2)^2
s = 26 - 9.8(4)
s = 26 - 39.2
s = -13.2
The negative value indicates that the ball is below the ground level. In this context, we're interested in the height above the ground, so we consider the magnitude of the displacement.
Therefore, the ball is 13.2 meters above the ground after 2 seconds.
Note: If we assume the ground level as the reference point, we can simply drop the negative sign and state that the ball is 13.2 meters high above the ground after 2 seconds.