)A ball is hit straight up with an initial speed of 50 meters per second. What is the speed of the ball 3 seconds after it is hit? [Neglect friction.]

The velocity of the ball after 3 seconds it is hit, is 20.6 m/s

Explanation:

Given;

initial velocity of the ball, u = 50 m/s

time of the upward motion of the ball, t = 3 seconds

The velocity of the ball after 3 seconds is calculated by applying the following kinematic equation;

v = u - gt

where;

g is acceleration due to gravity = 9.8 m/s²

v = 50 - (9.8 x 3)

v = 50 - 29.4

v = 20.6 m/s

Therefore, the velocity of the ball after 3 seconds it is hit, is 20.6 m/s

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

correct it is c.) b b