When the ball is thrown straight up, it follows a free fall motion, which is a uniformly accelerated motion with constant acceleration ( $g=9.8 m/s^2$ towards the ground). Therefore, we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

In this problem, we have:

u = 44.1 m/s is the initial vertical velocity of the ball

v = 0 is the final velocity when the ball reaches the maximum height

s is the maximum height

$a=-g=-9.8 m/s^2$ is the acceleration of gravity (downward, so negative)

Solving for s, we find the maximum height reached by the ball:

$s=-\frac{u^2}{2a}=-\frac{44.1^2}{2(-9.8)}=99.2 m$

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3 years ago

What is the direction of the field halfway between two horizontal parallel wires if the top wire has a current of 4 A to the lef

Alex777 [14]

Answer:

A) Out of the page.

Explanation:

Right-hand rule points the direction of the magnetic field at any point.

<u>Top wire</u>: Current is to the left. Point your thumb to the left and curl your other fingers around the wire. The tips of the four fingers points the direction of the field at that point. In this case, out of the page.

<u>Bottom wire</u>: Current is to the right. Point your thumb to the right and curl your other fingers around the wire. The tips of the four finger points out of the page again.

So, the total field produced by both wires is directed out of the page.

Another method to figure out the direction is the mathematical method.

Use the B-field formula:

$d\vec{B} = \frac{\mu_0}{4\pi}\frac{Id\vec{l}\times \^r}{r^2}$

The cross product between the direction of the current and the target position gives the direction of the B-field. If the left is -x direction and downwards is the -y direction, then

$(-\^x) \times (-\^y) = +\^z$ for the top wire.