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A second rock is thrown straight upward with a speed of 6.000 m/s. If this rock takes 1.443 s to fall to the ground, from what height h was it released?

User Arsonik
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1 Answer

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Final answer:

To determine the height at which the rock was released, we can use the equations of motion. Given the rock's initial velocity of 6.000 m/s and the time it takes to fall to the ground (1.443 s), we can solve for the height using the equation h = v0 * t - (1/2) * g * t^2. Plugging in the values, we find that the rock was released from a height of 5.245 meters.

Step-by-step explanation:

To calculate the height at which the rock was released, we can use the equations of motion. Since the rock is thrown straight upward, its initial vertical velocity is 6.000 m/s and it takes 1.443 s to fall back to the ground. The equation we can use is:

h = v0 * t - (1/2) * g * t^2

Where:

h is the height
v0 is the initial velocity (6.000 m/s)
t is the time (1.443 s)
g is the acceleration due to gravity (9.8 m/s^2)

Plugging in the values, we get:

h = (6.000 m/s) * (1.443 s) - (1/2) * (9.8 m/s^2) * (1.443 s)^2

Solving the equation gives us the height h = 5.245 m. Therefore, the rock was released from a height of 5.245 meters.

User Fredlahde
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