Solve Throw Up Problem: 15m, +12m/s & -12m/s

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Two rocks are thrown from a 15m cliff, one upward at +12m/s and the other downward at -12m/s, with air resistance ignored. The first rock's motion is described by the equation 15 = 12t + 0.5(9.8)t^2, which requires careful attention to sign conventions for accurate calculations. The direction of displacement and the acceleration due to gravity must be correctly identified to solve for the time and impact velocity of each rock. The discussion emphasizes the importance of consistent sign usage in kinematic equations. Correctly applying these principles will yield the correct time and velocity for both rocks upon impact.
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Homework Statement


Two rocks are thrown off the edge of a cliff that is 15m above the ground. The first rock is thrown upward, at a velocity of +12m/s. The second is thrown downward, at a velocity of -12.0m/s. Ignore air resistance. Determine how long it takes the first rock to hit the ground and at what velocity it hits. Determine how long it takes the second rock to hit the ground and at what velocity it hits.


Homework Equations





The Attempt at a Solution



15=12t+.5(9.8)t^2

To figure out time I tried v/g=t (12/9.8) but it was coming out saying the answer was wrong.
 
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nic0le said:
15=12t+.5(9.8)t^2

Is that for the first rock or the second?
 
That was for the first rock.
 
Check your sign convention. Take upwards to be positive and downwards negative. Then the velocity is positive, which you have correct. What is the net displacement-- up or down (i.e. should the 12m be positive or negative)? In which direction is the acceleration due to gravity (g) acting? If you get these correct, you should have the right solution.
 
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