Consider the position of a ball thrown down

with an initial speed of 18 m/s.
What will be its position after 2.8 s? Let
the initial position be 0. The acceleration of gravity is 9.8 m/s2.
Answer in units of m

To determine the position of the ball after 2.8 seconds, we can use the equation of motion for an object in free fall:

y = y₀ + v₀t + (1/2)at²

where:
y is the final position
y₀ is the initial position (0 in this case)
v₀ is the initial velocity (18 m/s in this case)
t is the time (2.8 seconds in this case)
a is the acceleration due to gravity (-9.8 m/s² in this case, considering downward direction as negative)

Now, let's substitute the given values into the equation:

y = 0 + (18)(2.8) + (1/2)(-9.8)(2.8)²

y = 0 + 50.4 + (-1/2)(9.8)(7.84)

y = 50.4 - 38.384

y = 12.016

Therefore, the position of the ball after 2.8 seconds will be approximately 12.016 meters below the initial position.