Question

description the midterm exam covers material from chapters 1 - 5. instructions from the list of choices, select the one best answer. multiple attempts not allowed. this test can only be taken once. force completion this test can be saved and resumed later. your answers are saved automatically. question completion status: moving to another question will save this response. question 32 question 32 of 57 2 points save a a ball is thrown vertically upward. the ball returns to the point of release after 3.5 s. what is the speed of the ball as it is released? 0 m/s 14 m/s 21 m/s 34 m/s 17 m/s

Explanation:

Step1: Analyze the motion time

The time of flight \(T = 3.5\ s\). For vertical - motion under gravity (neglecting air resistance), the time to reach the maximum height \(t\) is half of the time of flight. So \(t=\frac{T}{2}=\frac{3.5}{2}s = 1.75\ s\).

Step2: Use the kinematic equation

The kinematic equation \(v = v_0+at\). At the maximum height, \(v = 0\ m/s\), and \(a=-g=- 9.8\ m/s^{2}\) (taking the upward direction as positive).
From \(v = v_0+at\), we can solve for \(v_0\). Rearranging the equation gives \(v_0=-at\).
Substitute \(a=-9.8\ m/s^{2}\) and \(t = 1.75\ s\) into the equation: \(v_0=9.8\times1.75\ m/s\).
\(v_0 = 17.15\approx17\ m/s\)

Answer:

17 m/s