A stone is thrown vertically upward with a speed of 12.0 m/s. (a) How fast is it moving when it reaches a height of 5.0 m? (b) How long is required to reach this height?

To solve this problem, we can first use kinematic equations to find the time it takes for the stone to reach a height of 5.0 m. Then, we can use the same equations to find the velocity of the stone at that height.

The kinematic equation for displacement is:
y = (Vi * t) + (1/2 * a * t^2)

Where:
y = displacement or height (5.0 m)
Vi = initial velocity (12.0 m/s)
a = acceleration (acceleration due to gravity, approximately -9.8 m/s^2, considering upward motion as positive)
t = time

Now, let's solve for t:

5.0 = (12.0 * t) + (1/2 * -9.8 * t^2)
5.0 = 12.0t - 4.9t^2

To solve this quadratic equation, we can set it equal to zero and use the quadratic formula:

4.9t^2 - 12.0t + 5.0 = 0

Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

Where:
a = 4.9, b = -12.0, c = 5.0

t = (-(-12.0) ± √((-12.0)^2 - 4 * 4.9 * 5.0)) / (2 * 4.9)
t = (12.0 ± √(144.0 - 98.0)) / 9.8
t = (12.0 ± √46.0) / 9.8

Since time cannot be negative, we take the positive root:

t ≈ (12.0 + √46.0) / 9.8 ≈ 1.51 seconds

(a) Now we can find the velocity when the stone reaches a height of 5.0 m.
To find the velocity, we can use the equation:
Vf = Vi + (a * t)

Where:
Vf = final velocity (the velocity when the stone reaches a height of 5.0 m)
Vi = initial velocity (12.0 m/s)
a = acceleration (acceleration due to gravity, approximately -9.8 m/s^2)
t = time taken to reach the height (1.51 seconds)

Vf = 12.0 + (-9.8 * 1.51)
Vf ≈ 12.0 - 14.8
Vf ≈ -2.8 m/s (negative sign implies the stone is moving downward)

So, when the stone reaches a height of 5.0 m, it is moving with a velocity of approximately -2.8 m/s (downward).

(b) The time required to reach a height of 5.0 m is approximately 1.51 seconds.

To find the answers to these questions, we can use the kinematic equations of motion. We'll use the following equations:

1. v = u + at (Equation 1)
2. v^2 = u^2 + 2as (Equation 2)
3. s = ut + (1/2)at^2 (Equation 3)

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
s = displacement

First, let's find the time it takes for the stone to reach a height of 5.0 m.

Using Equation 3 (s = ut + (1/2)at^2):
5.0 m = (12.0 m/s)t + (1/2)(-9.8 m/s^2)t^2

Simplifying the equation:
4.9t^2 - 12t + 5 = 0

Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / 2a:
t = (-(-12) ± √((-12)^2 - 4 * 4.9 * 5)) / (2 * 4.9)
t ≈ 1.227 s (ignoring the negative solution)

So, it takes approximately 1.227 seconds for the stone to reach a height of 5.0 m.

Next, let's find the velocity of the stone when it reaches this height.

Using Equation 1 (v = u + at):
v = 12.0 m/s - 9.8 m/s^2 * 1.227 s
v ≈ 0.0 m/s

Therefore, the stone has a velocity of approximately 0.0 m/s when it reaches a height of 5.0 m.

In summary:
(a) The stone is not moving when it reaches a height of 5.0 m.
(b) It takes approximately 1.227 seconds to reach this height.

V^2 = Vo^2 + 2g*h

Vo = 12 m/s.
g = -9.8 m/s^2
h = 5 m.
V = ?

28.8 m/s