An arrow is launched upward with an initial speed of 100 meters per second (m/s). the equations above describe the constant-acceleration motion of the arrow, where v₀ is the initial speed of the arrow, v is the speed of the arrow as it is moving up in the air, h is the height of the arrow above the ground, t is the time elapsed since the arrow was projected upward, and g is the acceleration due to gravity (9.8 m/s²).
what is the maximum height from the ground the arrow will rise to the nearest meter?

510 m

Explanation:

The arrow is fired upward, so the motion of the arrow is a free fall motion, therefore we can find the maximum height by using the suvat equation

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have:

v = 0 (at the point of maximum height, the velocity of the arrow is zero)

u = 100 m/s (initial velocity)

 (acceleration of gravity, downward)

s = ? is the maximum height

Solving for s,  we find the maximum height of the arrow:

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

as technology advances, new experiments often expose problems in accepted theories. c

hope i

-lavira: )