Answer :
Final answer:
2 seconds after being thrown with an initial velocity of 11 m/s, the ball is 2 meters above the ground, calculated using the kinematic equation for position under constant acceleration.
Explanation:
To determine how high above the ground the ball is at 2.00 seconds after being thrown with an initial velocity of 11 m/s, we can use the kinematic equation for position under constant acceleration:
s = ut + ½ at²
where:
- s is the displacement (height in this case)
- u is the initial velocity (11 m/s upward)
- t is the time (2.00 s)
- a is the acceleration (using g = -10 m/s² downward)
Plugging the values into the equation, we get:
s = (11 m/s)(2.00 s) + ½(-10 m/s²)(2.00 s)²
s = 22 m - 20 m
= 2 m
Thus, the ball is 2 meters above the ground 2 seconds after it is thrown.
