Question

test information description 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 16 of 25 question 16 4 points (problem reference 3 - 2) a projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively. approximately how long does it take the projectile to reach the highest point in its trajectory? a. 4 s b. 8 s c. 2 s d. 16 s

Explanation:

Step1: Identify the relevant vertical - motion formula

The vertical - motion of a projectile is a free - fall motion. At the highest point of the trajectory, the vertical velocity \(v_y = 0\). The kinematic equation for vertical velocity is \(v_y=v_{0y}-gt\), where \(v_{0y}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity (\(g = 10m/s^{2}\)), and \(t\) is the time.
\[v_y = v_{0y}-gt\]

Step2: Solve for time \(t\)

We want to find the time \(t\) when \(v_y = 0\). Rearranging the equation \(v_y=v_{0y}-gt\) for \(t\), we get \(t=\frac{v_{0y}-v_y}{g}\). Given \(v_{0y}=40m/s\) and \(v_y = 0\), and \(g = 10m/s^{2}\), we substitute these values into the formula:
\[t=\frac{40 - 0}{10}=4s\]

Answer:

A. 4 s