6. What is the kinetic energy of the ball as it is half way through the fall?

(: Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about the amounts of potential and kinetic energy the bowling ball has:
• as it sits on top of a building that is 40 meters tall.
• as it is half way through a fall off a building that is 40 meters tall and travelling 19.8 meters per second.
• as it is just about to hit the ground from a fall off a building that is 40 meters tall and travelling 28 meters per second

Explanation:

It is given that halfway through the fall, the speed of the ball is

v' = 19.8 m/s

The kinetic energy of the ball is given by

where

m = 2 kg is the mass of the ball

v' = 19.8 m/s is its speed

Substituting, we find the kinetic energy when the ball is halfway through the fall:

KE=1/2(2)(19.8)^2392J

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

don't remove my answer im trying my best to answer them

explanation: