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: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 moving to another question will save this response. question 18 of 25 question 18 4 points save answer (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. determine the maximum height reached by the projectile. a. 80 m b. 160 m c. 320 m d. 240 m

Explanation:

Step1: Identify the relevant kinematic - equation

The kinematic equation for vertical motion $v_y^2 = v_{0y}^2-2gh$ is used. At the maximum - height, the vertical component of velocity $v_y = 0$. The initial vertical velocity $v_{0y}=40\ m/s$ and the acceleration due to gravity $g = 10\ m/s^2$.

Step2: Rearrange the equation to solve for height $h$

Starting from $v_y^2 = v_{0y}^2 - 2gh$, when $v_y = 0$, we can solve for $h$:
\[h=\frac{v_{0y}^2}{2g}\]
Substitute $v_{0y}=40\ m/s$ and $g = 10\ m/s^2$ into the formula:
\[h=\frac{40^2}{2\times10}=\frac{1600}{20}=80\ m\]

Answer:

A. 80 m