An athlete throws a ball straight upward with an initial speed of 12 meters per second. The ball

reaches a maximum height and then falls back down toward the athlete. What are the

magnitudes of the velocity and acceleration of the ball at its maximum height?

O A. velocity = 0 m/s; acceleration = 0 m/s

B. velocity = 0 m/s, acceleration = 9.8 m/s

O C. velocity = 12 m/s; acceleration = 0 m/s

OD. velocity = 12 m/s; acceleration = 9.8 m/s

B. velocity = 0 m/s, acceleration = 9.8 m/s

Explanation:

When the ball reaches the maximum height all energy is potential energy. Because of that, the kinetic energy of the ball is zero and the velocity is also zero. The acceleration of the ball is in all trajectory the gravitational constant, because the gravitational force always acts over the ball.

Hence, the velocity is 0m/s and the acceleration is 9.8m/s^2

B. velocity = 0 m/s, acceleration = 9.8 m/s

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west