To solve this problem, we can use the principles of projectile motion and apply the equations of motion for vertical motion.
Let's consider the following variables:
- u = initial velocity = 16.5 m/s (upward)
- v = final velocity = u/2 = 16.5/2 = 8.25 m/s (upward)
- a = acceleration = -9.8 m/s² (acceleration due to gravity, downward)
- s = displacement = ? (height above the initial position)
- t = time taken = ?
We can use the equation of motion:
v = u + at
Substituting the known values:
8.25 = 16.5 - 9.8t
Rearranging the equation:
9.8t = 16.5 - 8.25
9.8t = 8.25
t = 8.25 / 9.8
t ≈ 0.8418 seconds
Now, we can use the equation of motion to calculate the height:
s = ut + (1/2)at^2
Substituting the known values:
s = 16.5 * 0.8418 + (1/2) * (-9.8) * (0.8418)^2
s ≈ 13.85 meters
Therefore, the ball will be approximately 13.85 meters above its initial position when it has one-half its initial speed.