Asked by poi
a cannonaball is fired stright up into the air at a speed of 80 ft/sec from a height of 2ft. A. when will the cannon ball be 98 ft in the air?
B. when will the cannonball hit the ground?
B. when will the cannonball hit the ground?
Answers
Answered by
Damon
Vi = + 80
g = -32 ft/s^2 (old text, have not seen these units for years)
h = 2 + Vi t - (32/2) t^2
98 = 2 + 80 t - 16 t^2
16 t^2 - 80 t + 96 = 0
t^2 - 5 t + 6 = 0
(t-2)(t-3) = 0
t = 2 or 3 seconds
Two on the way up, three on the way down
When will h = 0 ?
0 = 2 + 80 t - 16 t^2
8 t^2 - 40 t -2 = 0
t = [ 40 +/- sqrt(1600 +64) ]/ 16
t = [ 40 +/- 40.8 ] / 16
negative time was before we pulled the trigger, use + solution
t = 80.8/16 = 5.05 seconds
g = -32 ft/s^2 (old text, have not seen these units for years)
h = 2 + Vi t - (32/2) t^2
98 = 2 + 80 t - 16 t^2
16 t^2 - 80 t + 96 = 0
t^2 - 5 t + 6 = 0
(t-2)(t-3) = 0
t = 2 or 3 seconds
Two on the way up, three on the way down
When will h = 0 ?
0 = 2 + 80 t - 16 t^2
8 t^2 - 40 t -2 = 0
t = [ 40 +/- sqrt(1600 +64) ]/ 16
t = [ 40 +/- 40.8 ] / 16
negative time was before we pulled the trigger, use + solution
t = 80.8/16 = 5.05 seconds
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.