A ball is thrown straight upward. At 4.36 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?

17.4

To find the maximum height above its launch point that the ball attains, we need to use the concept of projectile motion.

First, let's define the variables:
- h_max: Maximum height above the launch point
- v_launch: Launch speed of the ball
- v_final: Speed of the ball at 4.36 m above the launch point

We are given that at 4.36 m above its launch point, the ball's speed (v_final) is one-half of its launch speed (v_launch). Mathematically, we can express this as:

v_final = (1/2) * v_launch

We also know that at the maximum height, the vertical component of the ball's velocity is 0. This means that the ball has stopped momentarily and is just about to come back down. Using this information, we can set up the following equation:

v_final^2 = v_launch^2 + 2 * g * (h_max - 0)

where g is the acceleration due to gravity.

Since v_final is given as one-half of v_launch, we can substitute this value into the equation:

(v_launch/2)^2 = v_launch^2 + 2 * g * h_max

Now, we can solve for h_max:

(v_launch^2)/4 = v_launch^2 + 2 * g * h_max

Multiplying both sides by 4:

v_launch^2 = 4 * v_launch^2 + 8 * g * h_max

Rearranging the equation:

-3 * v_launch^2 = 8 * g * h_max

Simplifying:

h_max = (-3 * v_launch^2) / (8 * g)

Now, we can substitute the known values to find the maximum height:

h_max = (-3 * v_launch^2) / (8 * g)

Given that g is approximately 9.8 m/s^2, we can calculate the maximum height.