Asked by Sally
A sand bag is dropped out of a balloon traveling upward at speed of 2.3m/s. A kiddo drops a bag at the instant the balloon is 34m above the ground. How long does it take the sand bag to hit the ground? I got -2.41s why?
Answers
Answered by
Damon
h = 0
Hi = 34
Vi = + 2.3
(1/2) a = -g/2 = -4.9
h = Hi + Vi t - 4.9 t^2
0 = 34 + 2.3 t - 4.9 t^2
4.9 t^2 - 2.3 t - 34 = 0
t = [ 2.3 +/- sqrt(5.29 + 881) ]/ 9.8
t = [ 2.3 +/- 29.8 ]/9.8
t = 3.27 OR -2.8
the negative t was when it would have been at ground on the way up if it were thrown up from ground and is not relevant.
Hi = 34
Vi = + 2.3
(1/2) a = -g/2 = -4.9
h = Hi + Vi t - 4.9 t^2
0 = 34 + 2.3 t - 4.9 t^2
4.9 t^2 - 2.3 t - 34 = 0
t = [ 2.3 +/- sqrt(5.29 + 881) ]/ 9.8
t = [ 2.3 +/- 29.8 ]/9.8
t = 3.27 OR -2.8
the negative t was when it would have been at ground on the way up if it were thrown up from ground and is not relevant.
Answered by
Sally
I don't understand the negative. Would it be wrong if I put the negative answer if they're asking how long it takes the sand bag to hit the ground?
Answered by
Damon
Yes, the negative is unreal and wrong.
The trajectory is a parabola. It crosses the x axis twice. However the first crossing is in our imaginations. It is before the kid dropped the bag, as if the bag had been thrown up from the ground instead of starting 34 meters up with an initial speed up
The trajectory is a parabola. It crosses the x axis twice. However the first crossing is in our imaginations. It is before the kid dropped the bag, as if the bag had been thrown up from the ground instead of starting 34 meters up with an initial speed up
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