Asked by ana22
For a flourish at the end of her act, a juggler tosses a single ball high in the air. She catches the ball 3.6 s later at the same height from which it was thrown.
Part A-What was the initial upward speed of the ball?2sf
Answers
Answered by
maria
I think that the answer is 1.8
3.6/2(for the time it toke to go up then back down)=1.8
I am only in 7th grade so i am just guessing hope this is the right answer though :)
3.6/2(for the time it toke to go up then back down)=1.8
I am only in 7th grade so i am just guessing hope this is the right answer though :)
Answered by
Writeacher
Maria is guessing? Who would believe someone who admits she is guessing?
Answered by
Damon
reaches top 3.6/2 or 1.8 seconds after launch
at top, v = 0
but
v = Vi - g t
in meters, seconds g = 9.81 m/s^2
so
0 = Vi -9.81 t
0 = Vi -9.81*1.8
Vi = 17.7 meters/second
at top, v = 0
but
v = Vi - g t
in meters, seconds g = 9.81 m/s^2
so
0 = Vi -9.81 t
0 = Vi -9.81*1.8
Vi = 17.7 meters/second
Answered by
allie
if the weather channel guessed that it would rain, you would not believe them even if they have evidence of a thunder storm. My sister said that she assumed that the ball's speed would be the same going up and down so she gave "evidence" for why she thought the answer was 1.8
Answered by
Amanda
How did Damon get 9.81 meters/second?
Answered by
Dan
9.81 m/s is the constant for gravity, or acceleration during free fall. He is correct and his math works out as well.
Because you were only given the time it took for the ball to reach its initial position half of the time it spent in travel is against gravity and the other in free fall.
So the formula I used to get the answer was: Vi=(1/2t)*gravity(9.81m/s)
Hope this helped.
Because you were only given the time it took for the ball to reach its initial position half of the time it spent in travel is against gravity and the other in free fall.
So the formula I used to get the answer was: Vi=(1/2t)*gravity(9.81m/s)
Hope this helped.
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