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

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In summary, two rocks are thrown off a 15m cliff with velocities of +12m/s and -12.0m/s. Ignoring air resistance, the time it takes for the first rock to hit the ground and its velocity at impact can be determined by the equation 15=12t+.5(9.8)t^2. The time can be found using v/g=t with a positive velocity and acceleration due to gravity acting downwards. The same process can be used for the second rock.
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nic0le
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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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  • #2
nic0le said:
15=12t+.5(9.8)t^2

Is that for the first rock or the second?
 
  • #3
That was for the first rock.
 
  • #4
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.
 

1. How do you solve the throw up problem with given values of 15m, +12m/s, and -12m/s?

The throw up problem involves calculating the height of an object thrown vertically with an initial velocity of 12m/s and a final velocity of -12m/s, and a total distance of 15m. To solve this, we can use the formula h = (vf2 - vi2) / 2g, where h is the height, vi is the initial velocity, vf is the final velocity, and g is the acceleration due to gravity (9.8m/s2). Plugging in the values, we get h = (144 - 144) / 19.6 = 0m. Therefore, the height of the object at its peak is 0 meters.

2. What is the significance of the positive and negative velocities in the throw up problem?

In the throw up problem, the positive velocity (+12m/s) represents the initial upward velocity of the object, while the negative velocity (-12m/s) represents the final downward velocity of the object. This indicates that the object is thrown upwards, reaches its peak, and then falls back down due to gravity.

3. How is the distance of 15m related to the throw up problem?

The distance of 15m in the throw up problem represents the total vertical distance traveled by the object. This can be calculated by using the formula d = vit + 1/2at2, where d is the distance, vi is the initial velocity, t is the time, and a is the acceleration. Since we are only considering the vertical motion of the object, we can ignore the horizontal distance and use only the vertical distance of 15m to solve the problem.

4. Can the throw up problem be solved without knowing the initial and final velocities?

Yes, the throw up problem can be solved without knowing the initial and final velocities. If we know the height and the time it takes for the object to reach its peak, we can use the formula h = vit + 1/2at2 to calculate the initial velocity. Similarly, we can use the formula h = vft - 1/2at2 to calculate the final velocity. Therefore, the throw up problem can be solved using different combinations of known values.

5. How can the throw up problem be applied in real life scenarios?

The throw up problem can be used to determine the height of an object thrown vertically, such as a ball or a projectile, as well as other factors such as the initial and final velocities and the time it takes for the object to reach its peak. This problem can be applied in sports, such as basketball or baseball, to analyze the trajectory of a thrown ball. It can also be used in engineering to calculate the height of a building or the path of a rocket. Additionally, the throw up problem can help in understanding the concept of projectile motion and the effects of gravity on objects in motion.

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