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question 10 4 points save answer
the same arrow, from the previous question, is now dropped from the same height of 4.9 m above the ground. neglecting friction, how long will it take the arrow to hit the ground?
1.0 s
1.4 s
6.0 s
9.2 s
4.6 s

Explanation:

Step1: Identify the kinematic - equation

The arrow is in free - fall, and the kinematic equation for vertical displacement $y = v_0t+\frac{1}{2}gt^2$. Since it is dropped ($v_0 = 0$), the equation simplifies to $y=\frac{1}{2}gt^2$.

Step2: Solve for time $t$

We know $y = 4.9$ m and $g = 9.8$ m/s². Rearranging the equation $y=\frac{1}{2}gt^2$ for $t$, we get $t=\sqrt{\frac{2y}{g}}$.
Substitute $y = 4.9$ m and $g = 9.8$ m/s² into the formula: $t=\sqrt{\frac{2\times4.9}{9.8}}$.
First, calculate the value inside the square - root: $\frac{2\times4.9}{9.8}=\frac{9.8}{9.8}=1$. Then, $\sqrt{1}=1$ s.

Answer:

A. 1.0 s