Question

A volcano throws a stone that reaches a maximum height of 9220 meters directly up from earth. Neglecting air resistance, what is the initial velocity of the stone in the vertical direction?

Answer 1

The initial velocity of the stone in the vertical direction is approximately 429.4 m/s.

Step-by-step explanation:

To find the initial velocity of the stone in the vertical direction, we can use the equation for vertical motion:

vf^2 = vi^2 + 2ad

Since the stone reaches a maximum height directly up, the final velocity at the top of the path is 0 m/s. The initial velocity in the vertical direction (vi) is what we're trying to find. The acceleration (a) is equal to the acceleration due to gravity (-9.8 m/s^2). And the displacement (d) is the maximum height reached by the stone, which is 9220 meters.

Plugging in the values, we have:

0 = vi^2 + 2(-9.8)(9220)

Simplifying the equation, we get:

vi^2 = 2(9.8)(9220)

Taking the square root of both sides, we find:

vi = ± sqrt(2(9.8)(9220))

Calculating the value, we have:

vi ≈ ± 429.4 m/s

Since the stone is thrown upwards, the initial velocity in the vertical direction is positive. Therefore, the initial velocity of the stone in the vertical direction is about 429.4 m/s.