A ball is thrown straight upward and rises to a maximum height of 23 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^2 = Vo^2 + 2g*h

Vo^2 = V^2 - 2g*h.
Vo^2 = 0 + 19.6*23 = 450.8
Vo = 21.2 m.

h = (0.5Vo)^2-Vo^2)/2g
h = ((10.62)^2-(21.2^2))/-19.6=17.2 m.

To find the height above its launch point where the speed of the ball has decreased to one-half of its initial value, we can use the concept of projectile motion.

Step 1: Identify the relevant equations
For projectile motion in the vertical direction, the equation for the velocity at any point can be given as:

v(final) = v(initial) - g*t

where:
- v(final) is the final velocity
- v(initial) is the initial velocity
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time

Step 2: Determine the initial velocity
From the information given in the problem, the ball is thrown straight upward, meaning the initial velocity is positive because it's moving in the positive direction. Therefore, we have:

v(initial) = +v0

Step 3: Find the time it takes for the speed to decrease to half
We know that velocity is equal to speed, so we can rewrite the equation as:

v(final) = v(initial) - g*t
v(final) = (1/2)*v(initial)
(1/2)*v(initial) = v(initial) - g*t

Now we need to solve for t:

(1/2)*v(initial) = v(initial) - g*t
(1/2)*v(initial) - v(initial) = -g*t
-(1/2)*v(initial) = -g*t
t = (1/2)*v(initial)/g

Step 4: Calculate the height
We can use the equation for vertical displacement:

h = v(initial)*t - (1/2)*g*t^2

Substituting the value of t that we found in step 3, we get:

h = v(initial)*[(1/2)*v(initial)/g] - (1/2)*g*[(1/2)*v(initial)/g]^2

Simplifying:

h = (1/2)*[v(initial)]^2/g - (1/8)*[v(initial)]^2/g

Step 5: Solve for the height
Given that the maximum height reached is 23 m, we can set this equation equal to 23 and solve for the height where the speed decreases to one-half:

23 = (1/2)*[v(initial)]^2/g - (1/8)*[v(initial)]^2/g

Simplifying and solving for [v(initial)]^2/g:

23 = (4/8)*[v(initial)]^2/g - (1/8)*[v(initial)]^2/g
23 = (3/8)*[v(initial)]^2/g
[v(initial)]^2/g = (8/3)*23
[v(initial)]^2/g = 61.333
[v(initial)]^2 = 61.333*g
[v(initial)]^2 = 61.333*9.8
[v(initial)]^2 = 600.7934
[v(initial)] = sqrt(600.7934)
[v(initial)] = 24.5 m/s (approximately)

Now we can substitute this value back into the equation for height to find:

h = (1/2)*[v(initial)]^2/g - (1/8)*[v(initial)]^2/g
h = (1/2)*(24.5)^2/9.8 - (1/8)*(24.5)^2/9.8
h = 23.75 m (approximately)

Therefore, the height above its launch point where the speed of the ball has decreased to one-half of its initial value is approximately 23.75 m.

To determine the height above the launch point at which the speed of the ball decreases to one-half of its initial value, we can use the principles of kinematics.

The equation that relates the final velocity (vf), initial velocity (vi), acceleration (a), and displacement (d) of an object in one-dimensional motion is:

vf^2 = vi^2 + 2ad

In this case, the ball is thrown straight upward, so its acceleration is due to gravity, g, which is approximately -9.8 m/s^2 (negative due to being in the opposite direction of the ball's upward motion).

Given that the ball rises to a maximum height of 23 m, we know that the displacement (d) is 23 m.

At the maximum height, the ball momentarily comes to rest, so its final velocity (vf) would be 0 m/s.

Since we're looking for the height at which the speed of the ball decreases to one-half of its initial value, we need to find the initial velocity (vi) at that point. It will be half of the original initial velocity.

Now, we can rearrange the above equation to solve for the displacement (d):

vf^2 = vi^2 + 2ad
0 = vi^2 + 2(-9.8)(23)
0 = vi^2 - 448.4

Solving for vi^2:

vi^2 = 448.4

Taking the square root of both sides:

vi ≈ 21.17 m/s

Therefore, the initial velocity (vi) at the height where the speed is one-half of its initial value is approximately 21.17 m/s.

Now, we can use the equation for the final velocity in free fall to determine the height (h) at which the speed decreases to one-half of its initial value:

vf = vi + gt

Rearranging the equation, we have:

t = (vf - vi) / g

Substituting the known values:

t = (0 - 21.17) / -9.8

t ≈ 2.16 s

Now, we can use the equation for displacement in free fall:

d = vi*t + 0.5*g*t^2

Substituting the known values:

d = 21.17 * 2.16 + 0.5*(-9.8)*(2.16)^2

d ≈ 24.5 m

Therefore, the height above its launch point at which the speed of the ball decreases to one-half of its initial value is approximately 24.5 m.

m g h = (1/2)m Vi^2

9.81 (23) = (1/2) Vi^2
so
Vi = 21.2 m/s initial speed up

v = Vi - g t
10.6 = 21.2 - 9.81 t
t = 1.08 seconds

h = 0 + Vi t - (1/2)g t^2
= 21.2(1.08) - 4.9 (1.08)^2
= 17.2 meters