A ball is thrown straight upward. At 6.30 m above its launch point, the ball’s speed is one-half its launch speed. What maximum height above its launch point does the ball attain?

To determine the maximum height above its launch point that the ball attains, we can use the following steps:

Step 1: Identify the given information.
The problem provides the following information:
- At a height of 6.30 m above its launch point, the ball's speed is one-half its launch speed.

Step 2: Find relevant equations.
We can use equations of motion to solve this problem. The key equations we need are:
- Equation 1: v² = u² + 2as (equation of motion relating initial velocity (u), final velocity (v), acceleration (a), and displacement (s))
- Equation 2: v = u + at (equation of motion relating initial velocity (u), final velocity (v), acceleration (a), and time (t))

Step 3: Applying the equations.
Since the ball is thrown straight upward, the acceleration due to gravity will act in the opposite direction to the initial velocity (downward).

Let's assign the following variables:
u = Initial velocity (launch speed)
v = Final velocity (half the launch speed as given)
s = Maximum height above the launch point
a = Acceleration due to gravity (-9.8 m/s²)

Using Equation 1, we can write:
(v/2)² = u² + 2as
Simplifying:
(v²)/4 = u² + 2as

Using Equation 2, we know that the final velocity when the ball reaches its maximum height is 0 m/s (ball momentarily stops). Therefore:
0 = u + at

Step 4: Solving for the unknowns.
We now have two equations with two unknowns (u and s). We can rearrange the second equation to solve for the time taken (t):
0 = u + at
t = -u/a

Substituting this value of t into the first equation:
(v²)/4 = u² + 2a(-u/a)s
(v²)/4 = u² - 2us
(v²)/4 = u(u - 2s)

Rearranging this equation:
v² = 4u(u - 2s)
v² = 4u² - 8us

Substituting the value of v (v = u/2):
(u/2)² = 4u² - 8us
(u²)/4 = 4u² - 8us
(u²)/4 - 4u² + 8us = 0
-3(u²)/4 + 8us = 0
8us = 3u²/4

Simplifying:
8s = 3u/4
s = (3/32)u

Now we know that s is equal to (3/32)u.

Step 5: Substitute values and calculate.
To find the maximum height, we need to determine the value of s using the given information that the speed at 6.30 m above the launch point is half the launch speed (u/2).

Plugging this value into the equation, we have:
6.30 = (3/32)u

To solve for u, we rearrange the equation:
u = (6.30 * 32) / 3
u = 211.2 m/s

Now, substitute this value back into the equation for s:
s = (3/32) * 211.2
s ≈ 19.8 m

Therefore, the maximum height above the launch point that the ball attains is approximately 19.8 meters.