Asked by Anonymous
two balls are dropped in a vacuum. if one is caught in half the time as the other what is the difference in distance the balls traveled?
a.1/4*t
b.1/4*t^2
c.1/2*t^2
d.15/3*t
e.15/4*t^2
I believe it is b. but i'm not 100% sure.
a.1/4*t
b.1/4*t^2
c.1/2*t^2
d.15/3*t
e.15/4*t^2
I believe it is b. but i'm not 100% sure.
Answers
Answered by
bobpursley
ball1: d=1/2 g t^2
ball2: d=1/2 g (t/2)^2
difference in distance
ball1-ball2=1/2 gt^2 (1-1/4)=
ball1(3/4)
ball2 went 1/4 as far as ball1.
The difference in distance is 3/4 the distance ball1 went.
I am not certain your teacher really meant "difference'.
ball2: d=1/2 g (t/2)^2
difference in distance
ball1-ball2=1/2 gt^2 (1-1/4)=
ball1(3/4)
ball2 went 1/4 as far as ball1.
The difference in distance is 3/4 the distance ball1 went.
I am not certain your teacher really meant "difference'.
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.