Aball is thrown vertically downward from the top of a 30.6-m-tall building. the ball passes the top of a window that is 10.7 m above the ground 2.00 s after being thrown. what is the speed of the ball as it passes the top of the window?

v = 19.6 m/s

Explanation:

Height of building = 30.6 m.

Height of window from the ground level= 10.7 m.

Acceleration due to gravity = 9.8

At initial condition ball at rest condition so u= 0 m/s.

Lets take when passes through the window ,velocity is v.

Here acceleration is constant so we can apply motion equation .

We know that

v= u + a t

So by putting the values

v = 0 +9.8 x 2

v = 19.6 m/s

So the velocity of ball is 19.6 m/s when passes through the window after 2 s.

it is very well related to a lot of things but with my limited knowledge i can confidently say that the answer is space.

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: