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.

multiply 2kg by latent heat of vap

answer; how far out of place which the object finished relative to the starting location;