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Old Sep 1st 2018, 02:56 PM   #1
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Need help with this question regarding kinematics.

a) A ball is thrown in the -y direction off of a cliff with a velocity of 7m/s. If the ball takes 1.45s to reach the ground, how high off of the ground is the cliff? (Answer: -20m)

Im not getting -20m for some reason when i plug in the numbers.

Known: A=-9.8m/s^2
T=1.45s
Initial velocity= 0m/s
Displacement is unknown
I suspect that the final velocity is -7m/s
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Old Sep 1st 2018, 03:08 PM   #2
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Why do you say that the initial velocity is 0, when you are clearly told it's -7 m/s. You aren't asked about the final velocity, so don't be concerned about it. Use the standard equation for distance based on time, initial displacement, initial velocity, and acceleration:

$\displaystyle y(t) = y_0 + v_o t + \frac 1 2 a t^2$

Here y(1.45s) = 0 (because the ball hits the ground at t=1.45s), v_0 = -7 m/s (negative because the ball is thrown downward from the top of the cliff), a = -9.8 m/s^2, and t = 1.45 s. Plug and chug to solve for y(1.45). By the way, the answer is not negative 20 meters - that would mean the cliff is actually a well. The answer should be a positive number.
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Old Sep 1st 2018, 04:59 PM   #3
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Originally Posted by Palilth View Post
Known: A=-9.8m/s^2
T=1.45s
Just an FYI: Physics is "case-sensitive." A and a are different variables.

Typically the acceleration due to gravity has the symbol "g." If you want to call it nothing more than an acceleration that's fine, but then the symbol is "a." In Kinematics time is usually a "t" but if you have several times "T" can be used as well.

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