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 9 of 25 question 9 4 points save answer an arrow is shot horizontally from a height of 4.9 m above the ground. the initial speed of the arrow is 45 m/s. neglecting friction, how long will it take the arrow to hit the ground? 1.4 s 9.2 s 4.6 s 6.0 s 1.0 s

Explanation:

Step1: Identify vertical - motion equation

The vertical - displacement of the arrow is given by the equation $y = y_0+v_{0y}t-\frac{1}{2}gt^2$. Since the arrow is shot horizontally, $v_{0y} = 0$. The initial height $y_0=4.9$ m and when it hits the ground $y = 0$. So the equation becomes $0=y_0-\frac{1}{2}gt^2$.

Step2: Solve for time $t$

Rearranging the equation $0 = y_0-\frac{1}{2}gt^2$ for $t$, we get $t=\sqrt{\frac{2y_0}{g}}$. Given $y_0 = 4.9$ m and $g = 9.8$ m/s², then $t=\sqrt{\frac{2\times4.9}{9.8}}$.

Step3: Calculate the value of $t$

$t=\sqrt{\frac{9.8}{9.8}}=1.0$ s

Answer:

E. 1.0 s