Question

this final exam is cumulative and covers material from the entire course.
instructions: from the list of choices, select the one best answer.
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question 15 of 62
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question 15
1.6 points
a dynamite blast at a quarry launches a chunk of rock straight upward, and 3.0 s later, it is rising at a speed of 12 m/s. what was the initial speed of the rock?
hint: remember your vector directions, call vectors directed upward + and those directed downward are -.
○ 52 m/s
○ 28 m/s
○ 41 m/s
○ 34 m/s

Explanation:

Step1: Identify the kinematic equation

We use the kinematic equation \( v = v_0 + at \), where \( v \) is the final velocity, \( v_0 \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. For vertical motion, the acceleration due to gravity \( a=-g=-9.8\ m/s^2 \) (negative because it's downward), \( v = 12\ m/s \) (upward, so positive), and \( t = 3.0\ s \).

Step2: Rearrange the equation to solve for \( v_0 \)

Rearranging \( v = v_0 + at \) gives \( v_0=v - at \).

Step3: Substitute the values

Substitute \( v = 12\ m/s \), \( a=-9.8\ m/s^2 \), and \( t = 3.0\ s \) into the equation:
\( v_0=12-( - 9.8\times3.0)=12 + 29.4 = 41.4\approx41\ m/s \)

Answer:

41 m/s (corresponding to the option "41 m/s")