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

[louis676]

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 =?
X_o = 79 m
V_o = 0 m/s
A = -9.8 m/s ²

X = X_o + V_o 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) ²Position of elevating object
x =?
Δt =?
X_o = 0 m
V_o = 28 m/s
A = -9.8 m/s ²

X = X_o + V_o 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.

 

[gneill]

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

 

[louis676]

So my answer is correct, right?

 

[gneill]

So my answer is correct, right?

I believe that is what I stated...

 

[louis676]

Alrite thank you so much.

 

[gneill]

Alrite thank you so much.

No problem. Cheers