For an uniformly accelerated motion, we can write

where

is the acceleration of this motion, which in this problem is the gravitational acceleration, with a negative sign because it points downward, against the direction of the motion; h=0.540 m is the distance covered by the flea, and

is the initial velocity.
At the maximum height, the velocity is zero, so

. Therefore we can solve to find

:
Answer:
C
Explanation:
Sound waves speed up noticeably when moving through a solid or liquid, because all it is is just particles colliding; and particles are way closer together with those states of matter.
The speed of light can change when moving through different substances, but this is dependent on complicated factors such as frequency, polarization, intensity, et. cetera
The important part is that it does change speed, so your answer is C.
Hope this helps!
Answer:

Explanation:
q = Charge
r = Distance




The electric field is given by

The electric field at the aircraft is 