a stone is thrown upward with a speed of 24.0m/s how fast is it moving when it reaches a height of 13.0m?

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

Vo = 24 m/s
g = -9.8 m/s^2
h = 13 m.
Solve for V.

To solve this problem, we can use the equations of motion. Specifically, we can use the equation for the final velocity of an object in free fall when it reaches a certain height:

v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity
a = acceleration (due to gravity, approximately -9.8 m/s²)
s = displacement

Given:
u = 24.0 m/s (upward)
s = 13.0 m
a = -9.8 m/s²

Let's substitute these values into the equation and solve for v:

v^2 = (24.0)^2 + 2(-9.8)(13.0)
v^2 = 576 + 2(-9.8)(13.0)
v^2 = 576 - 254.8
v^2 = 321.2

Taking the square root of both sides to find v:

v = √321.2
v ≈ 17.91 m/s

Therefore, the stone is moving at a speed of approximately 17.91 m/s when it reaches a height of 13.0 m.

To find the speed of the stone when it reaches a height of 13.0m, you need to use the concepts of kinematic equations. Specifically, you can use the equation that relates the initial velocity (u), final velocity (v), acceleration (a), and displacement (s):

v^2 = u^2 + 2as

In this case, the stone is thrown upward, so the initial velocity is 24.0 m/s (u = 24.0 m/s) and it reaches a height of 13.0m (s = 13.0m). The acceleration due to gravity can be considered as -9.8 m/s^2 (a = -9.8 m/s^2) since the stone is being thrown against gravity.

Now, let's plug in the known values into the equation:

v^2 = (24.0 m/s)^2 + 2 (-9.8 m/s^2)(13.0 m)

v^2 = 576.0 m^2/s^2 - 254.8 m^2/s^2

v^2 = 321.2 m^2/s^2

To find the velocity, take the square root of both sides:

v = √(321.2 m^2/s^2)

v ≈ 17.93 m/s

Therefore, the stone is moving at approximately 17.93 m/s when it reaches a height of 13.0 m.

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