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You throw a 3.00 N rock vertically into the air from ground level. You observe that when it is 13.0 m above the ground, it is traveling at 25.0 m/s upward. What is the initial velocity of the rock when it was thrown?

User Bravo
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Final answer:

To find the initial velocity of the rock, we use the kinematic equation for velocity involving initial velocity, acceleration, and displacement, which results in an initial velocity of approximately 29.66 m/s upward.

Step-by-step explanation:

The question presented involves finding the initial velocity of a rock thrown vertically into the air. Since the rock is subject to gravitational acceleration, we can use the kinematic equations for uniformly accelerated motion. Given that the rock's weight is 3.00 N and it is observed at a height of 13.0 m with an upward velocity of 25.0 m/s, we can use the equation for velocity as a function of initial velocity, acceleration, and displacement:

v2 = v02 + 2ad

Where v is the final velocity (25.0 m/s), v0 is the initial velocity (unknown), a is the acceleration due to gravity (-9.8m/s2 since it's upward), and d is the displacement (13.0 m). We can solve this equation for v0 (initial velocity).

Plugging in the values and solving for v0, we get:

25.02 = v02 - 2(9.8)(13.0)

625 = v02 - 254.8

v02 = 625 + 254.8

v02 = 879.8

v0 = √879.8

v0 = 29.66 m/s (approx.)

Therefore, the initial velocity of the rock was approximately 29.66 m/s upward.

User Blawzoo
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