How to find the time of two object passing each at opposite direction.

  • Thread starter louis676
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  • #1
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1. A ball is dropped from a height of 79 m. Simultaneously a ball is thrown upward from the ground at a velocity of 28 m/s. How much time passes until the balls are at the same height.

I need someone to check if my approach is correct.



Position of falling object
x =?
Δt =?
Xo = 79 m
Vo = 0 m/s
A = -9.8 m/s 2

X = Xo + Vo Δt + .5a (Δt) 2
X = 79 m + 0m/s (Δt) + .5(-9.8 m/s 2) (Δt) 2
X = 79 m - 4.9 m/s 2(Δt) 2


Position of elevating object
x =?
Δt =?
Xo = 0 m
Vo = 28 m/s
A = -9.8 m/s 2

X = Xo + Vo Δt + .5a (Δt) 2
X (Δt) = 0m + 28m/s Δt + .5a (Δt) 2
X (Δt) = 28 m/s (Δt) + .5(-9.8 m/s 2) (Δt) 2
X (Δt) = 28 m/s (Δt) – 4.9 m/s 2 (Δt) 2

To find the time, we need to set both the equation equal to each other.
79 m – 4.9 m/s 2 (Δt) 2 = 28 m/s (Δt) - 4.9 m/s 2(Δt) 2
79m = 28 m/s (Δt)
2.8 s = Δt

Then I replace 2.8 for t in both of the equation and check if my time is correct

X = 79 m – 4.9 m/s 2 (Δt)2
X = 79 m – 4.9 m/s 2 (2.8 s)2
X = 79 m - 4.9 m/s 2 (7.84s 2)
X = 79 m - 38.416 m
X = 40.6 m

X = 28 m/s (Δt) – 4.9 m/s 2 (Δt) 2
X = 28 m/s (2.8s) - 4.9 m/s 2 (2.8 s)2
X = 78.4 m - 38.416 m
X = 40 m
The answer is approximately the same.
 

Answers and Replies

  • #2
gneill
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Your approach is fine, and the result of 2.8s is good (at least to one decimal).

Keeping a few more decimal places in intermediate results would make your check of the answer more convincing :wink:
 
  • #3
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So my answer is correct, right?
 
  • #4
gneill
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20,803
2,784
So my answer is correct, right?
I believe that is what I stated...
 
  • #5
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Alrite thank you so much.
 
  • #6
gneill
Mentor
20,803
2,784

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