Question

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description\tshow all you work, including units, on separate paper. follow the \problem solving method\.
\tthis final exam is cumulative and covers material from the entire course.
instructions\tfrom the list of choices, select the one best answer.
multiple attempts\tnot allowed. this test can only be taken once.
force completion\tthis test can be saved and resumed later.
\tyour answers are saved automatically.
question completion status:
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question 16
1.6 points save answer
a pellet is fired straight upwards with an initial speed of 24 m/s. what is the maximum height reached by the pellet?
○ 29 m
○ 19 m
○ 15 m
○ 58 m

Explanation:

Step1: Identify the formula for maximum height

For an object in vertical motion (fired upwards), the maximum height \( h \) reached can be found using the kinematic equation \( v_f^2 = v_i^2 + 2a\Delta y \), where \( v_f \) is the final velocity (0 at maximum height), \( v_i \) is the initial velocity, \( a \) is the acceleration (here, \( a=-g=-9.8\ m/s^2 \) as it's against gravity), and \( \Delta y = h \) (the maximum height). Rearranging for \( h \), we get \( h=\frac{v_f^2 - v_i^2}{2a} \). Since \( v_f = 0 \), this simplifies to \( h=\frac{-v_i^2}{2(-g)}=\frac{v_i^2}{2g} \).

Step2: Plug in the values

Given \( v_i = 24\ m/s \) and \( g = 9.8\ m/s^2 \), substitute into the formula:
\( h=\frac{(24)^2}{2\times9.8}=\frac{576}{19.6}\approx29.39\ m \), which is approximately \( 29\ m \).

Answer:

29 m (the option corresponding to 29 m, e.g., if the first option is "29 m", then the answer is the option with "29 m")