Dragonite
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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