Free fall problem involving two objects

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SUMMARY

The discussion revolves around a physics problem involving two stones thrown vertically, one from a 110 ft tower with an initial velocity of 60 ft/s and the other from the ground with an initial velocity of 85 ft/s. The correct time for both stones to reach the same height is determined to be 4.42 seconds, with the height being 18.73 meters. The solutions presented by the participants highlight the importance of accurately applying kinematic equations and unit conversions, with the classmate's solution being validated by the teacher.

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  • Understanding of kinematic equations, specifically Vf = Vi + gt and d = Vit - 0.5gt²
  • Ability to perform unit conversions between feet and meters
  • Knowledge of projectile motion principles
  • Basic algebra skills for solving equations
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  • Study the application of kinematic equations in vertical motion problems
  • Learn about the effects of gravity on projectile motion
  • Explore common pitfalls in unit conversion and rounding errors
  • Practice solving similar problems involving multiple objects in motion
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Homework Statement


A stone is thrown vertically into the air from a tower 110 ft. high at the same time that a second stone is thrown upward from the ground. The initial velocity of the first stone is 60 ft/s and that of the second stone 85 ft/s. When and where will the stones be at the same height from the ground?

*note* When I asked my teacher, he said the "thrown vertically" meant that the stone was thrown upward


Homework Equations


Vf = Vi+gt
d=Vit-0.5gt2


The Attempt at a Solution



I have two solutions. One by me and the other from my classmate. According to the teacher, her solution is correct, but I disagree.

*Solution of classmate*

Conversion of units:

110ft = 33.53m
60ft/s = 18.29m/s
85ft/s = 25.91m/s

d1+33.53m=d2
18.29t - 4.9t2 + 33.53 = 25.91t - 4.9t2
33.53 = 25/91t - 18.29t
t = 4.4s
18.29 (4.4) - 4.9 (4.4)2 + 33.53 = 19.13m

*My solution* - I believe that my classmate failed to take into account the first stone being thrown upwards. However, there is also something wrong with my solution.

Conversion of units:

110ft = 33.53m
60ft/s = 18.29m/s
85ft/s = 25.91m/s

18.29 = 9.8t
1.87 = t
18.29(1.87) - 0.5(9.8)(1.87)2 + 33.53 = d of stone 1 at 1.87s from the ground
d = 50.6m

25.91 = 9.8t
2.64 = t
25.91(2.64) - 0.5(9.8)(2.64)2 = d of stone 2 at 2.64s from the ground
d = 34.25

2.64 - 1.87 = 0.77s
9.8(0.77) = V
d = 2.91m
50.6 - 2.91 = 47.69m = d of stone 1 at 2.64s from the ground

47.69 - (7.55t + 0.5(9.8)t2) = 34.25 - (0.5(9.8)t2
13.44 = 7.55t
t = 1.78s
Plugging in t to the formula for t, d = 28.96m
47.69 - 28.96 = 18.73 m
1.78 + 2.64 = 4.42s

Therefore, my answer is 4.42 s at 18.73 m from the ground
 
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The solution of your classmate is correct. Your piecewise solution might be correct, but unnecessarily complicated with a lot of rounding errors. That can be the reason of the different results.

ehild
 
ehild said:
The solution of your classmate is correct. Your piecewise solution might be correct, but unnecessarily complicated with a lot of rounding errors. That can be the reason of the different results.

ehild

I see. Thank you very much :)
 

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