That's the cool thing about free fall. The amount of time it takes to fall remains the same.
In this case, a ball that is simply dropped from rest will fall at the same rate as a ball that had some umph in the horizontal direction.
Answer:
I hope it's helpful for you ☺️
Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e
q/r
2
, so
E=
(5.11×10
−11
m)
2
(8.99×10
9
N.m
2
/C
2
)(1.60×10
−19
C)
=5.51×10
11
N/C∼10
1
2N/C
making (e) the best choice for this question.
The solution would be like
this for this specific problem:
V^2 = 2AS = 2FS/M
V = sqrt(2FS/M) =
sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps
So the speed of the arrow as it leaves the bow
is 42.5 mps.
I am hoping that this answer has
satisfied your query and it will be able to help you in your endeavor, and if
you would like, feel free to ask another question.