Asked by Lisa K.
A water balloon is thrown horizontally at a speed of 2.00 m/s from a roof of a building that is 6.00 m above ground. At the same instant the balloon is released a second balloon is thrown straight down at 2.00 m/s from the same height. Determine which balloon hits the ground first and how much soon it hits the ground than the other balloon?
Answers
Answered by
Henry
First Baloon
h = Vo*t + 0.5g*t^2 = 6 m.
0 + 4.9t^2 = 6
t^2 = 1.22
Tf = 1.11 s. = Fall time.
Second Baloon
2*t + 4.9t^2 = 6
4.9t^2 + 2t - 6 = 0.
Use Quadratic Formula:
Tf = 0.921 s. = Fall time.
1.11 - 0.921 = 0.189 s.
The Fall time of 2nd baloon was 0.189 s
shorter.
h = Vo*t + 0.5g*t^2 = 6 m.
0 + 4.9t^2 = 6
t^2 = 1.22
Tf = 1.11 s. = Fall time.
Second Baloon
2*t + 4.9t^2 = 6
4.9t^2 + 2t - 6 = 0.
Use Quadratic Formula:
Tf = 0.921 s. = Fall time.
1.11 - 0.921 = 0.189 s.
The Fall time of 2nd baloon was 0.189 s
shorter.
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