Question

a large boulder is ejected vertically upward from a volcano with an initial speed of 40.6 m/s. air resistance may be ignored.
part a
at what time after being ejected is the boulder moving at a speed 21.0 m/s upward?
express your answer in seconds.
$t = $
part b
at what time is it moving at a speed 21.0 m/s downward?
express your answer in seconds.

Explanation:

Step1: Recall vertical velocity formula

The velocity of an object in vertical motion is given by $v = v_0 - gt$, where $v_0$ is initial upward velocity, $g=9.8\ \text{m/s}^2$ is gravitational acceleration, and $t$ is time.
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Part A: Upward 21.0 m/s

Step2: Set values for upward velocity

We have $v=21.0\ \text{m/s}$, $v_0=40.6\ \text{m/s}$. Rearrange for $t$:
$t = \frac{v_0 - v}{g}$

Step3: Calculate time for Part A

Substitute values:
$t = \frac{40.6 - 21.0}{9.8} = \frac{19.6}{9.8} = 2.0\ \text{s}$
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Part B: Downward 21.0 m/s

Step4: Set values for downward velocity

Downward velocity is negative: $v=-21.0\ \text{m/s}$. Use the same formula:
$t = \frac{v_0 - v}{g}$

Step5: Calculate time for Part B

Substitute values:
$t = \frac{40.6 - (-21.0)}{9.8} = \frac{61.6}{9.8} = 6.3\ \text{s}$

Answer:

Part A: $2.0$ s
Part B: $6.3$ s