Final answer:
After 2 seconds, the velocity of the ball thrown upwards with an initial velocity of 14 m/s is -5.62 m/s, indicating it is moving downwards. The ball's height above the ground after 2 seconds is 8.38 meters.
Step-by-step explanation:
The question involves the concepts of kinematics, specifically the equations of motion under the influence of gravity. In this scenario, gravity acts as a constant acceleration downwards at 9.81 m/s2, which will change the ball's velocity and position over time.
To determine the ball's velocity after 2 seconds when thrown upwards with an initial velocity of 14 m/s, we can use the following kinematic equation:
v = u + at
Where:
v = final velocity
u = initial velocity (14 m/s)
a = acceleration due to gravity (-9.81 m/s2, negative because it's in the opposite direction of the throw)
t = time (2 seconds)
Using the equation:
v = 14 m/s + (-9.81 m/s2)(2 s) = 14 m/s - 19.62 m/s = -5.62 m/s.
The negative sign indicates that the ball is moving downwards after 2 seconds.
For the ball's height after 2 seconds, we use another kinematic equation:
s = ut + rac{1}{2}at2
Where s is the displacement.
Plugging in the values:
s = (14 m/s)(2 s) + rac{1}{2}(-9.81 m/s2)(2 s)2
s = 28 m - 19.62 m
s = 8.38 m
So, the ball is 8.38 meters above the ground after 2 seconds.