a) The downward acceleration due to gravity is approximately 10 m/s². Using the standard coordinate system, +y points upward. We choose y = 0 at the bottom of the cliff, and note that because the phone is dropped, the initial velocity along the y-axis is zero. So, the kinematic equation in the vertical direction is given by y(t) = -gt² + 180 m. We want to figure out how long it will take to get to the ground, so we need to solve for tground where y(tground) = 0. Using y(t), we have that |2(180 m) tground = Evaluating, we find that tground = 6 sec.

a) The downward acceleration due to gravity is approximately 10 m/s². Using the standard coordinate system, +y points upward. We choose y = 0 at the bottom of the cliff, and note that because the phone is dropped, the initial velocity along the y-axis is zero. So, the kinematic equation in the vertical direction is given by y(t) = -gt² + 180 m. We want to figure out how long it will take to get to the ground, so we need to solve for tground where y(tground) = 0. Using y(t), we have that |2(180 m) tground = Evaluating, we find that tground = 6 sec.

Name: Can you show the algebra steps.. specifically how we got the "2". **No
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Description: Can you show the algebra steps.. specifically how we got the "2". **Note question is already answered just need explanation. a) The downward acceleration due to gravity is approximately 10 m/s². Usingthe standard coordinate system, +y points upward.

University Physics Volume 1

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Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter4: Motion In Two And Three Dimensions

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Can you show the algebra steps.. specifically how we got the "2". **Note question is already answered just need explanation.

a) The downward acceleration due to gravity is approximately 10 m/s². Using
the standard coordinate system, +y points upward. We choose y = 0 at the
bottom of the cliff, and note that because the phone is dropped, the initial
velocity along the y-axis is zero. So, the kinematic equation in the vertical
direction is given by
y(t) = -gt² + 180 m.
We want to figure out how long it will take to get to the ground, so we need to
solve for tground where
y(tground) = 0.
Using y(t), we have that
|2(180 m)
tground =
Evaluating, we find that tground = 6 sec.

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