A ball is thrown straight upward and rises to a maximum height of 20 m above its launch point. At what height above its launch point has the speed of the ball decreased to one-half of its initial value?

v = Vi - g t

at max height, v = 0
0 = Vi -gt
so t at top = Vi/g

How high does it go?
20 = Vi t - (1/2)g t^2
20 = Vi (Vi/g) - (1/2) gVi^2/g^2

20 = (1/2) Vi^2/g
40 (9.81) = Vi^2
Vi = 19.8 m/s

when does v = 19.8/2 or 9.9 m/s
v = Vi - g t
9.9 = 19.8 - 9.81 t
9.81 t = 9.9
t = 1 second

To find the height above the launch point at which the speed of the ball decreases to one-half its initial value, we can use the concept of energy conservation.

Let's break down the problem step by step:

Step 1: Determine the initial speed of the ball.

Given that the ball is thrown straight upward, we can assume that the initial speed is the same as the final speed when the ball reaches its maximum height.

Step 2: Calculate the initial speed of the ball.

Given the limited information in the question, we'll have to make some assumptions. Assuming no air resistance, we can use the principle of conservation of energy.

At the maximum height of 20 m above the launch point, the ball has zero kinetic energy (since it momentarily pauses before falling back down) and only potential energy due to its position. The potential energy at this point is given by mgh, where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height (20 m in this case).

Therefore, at the maximum height, the initial kinetic energy is equal to the potential energy:

(1/2)mv^2 = mgh

Simplifying the equation, we find:

v^2 = 2gh

Taking the square root of both sides:

v = √(2gh)

Since we are interested in finding when the speed decreases to one-half its initial value, we can solve for v/2:

(v/2)^2 = 2gh

Simplifying further, we get:

v^2/4 = 2gh

v^2 = 8gh

Step 3: Determine the height at which the speed decreases to one-half its initial value.

To find the height at which the speed becomes half of its initial value, we can equate the expression for v^2 we derived above to (v/2)^2:

8gh = v^2/4

Simplifying, we get:

8gh = v^2/4
32gh = v^2

Multiplying both sides by h:

32gh^2 = v^2h

Now, we need to substitute the equation for v^2 we derived earlier:

32gh^2 = 8gh

Simplifying further, we find:

4h = 20

h = 5

Therefore, the height above the launch point at which the speed of the ball decreases to one-half its initial value is 5 meters.