Aball tossed vertically upward from the ground next to a building passes the bottom of a window 1.8 s after being tossed and passes the top of the window 0.20 s later. the window is 2.0 m high from top to bottom. what was the ball's initial velocity? the unit vector j^ is directed upward. how far is the bottom of the window from the launch position? how high does the ball rise above the launch position?

The ball takes 0.2 sec to travel the height of the window, which is 2 m.

 Then the speed of the ball at that time was

We know that:

Where is the initial velocity.

So:

Then we have the equation for the position as a function of time.

Where

= initial position

r = position as a function of time

= initial velocity

g = acceleration of gravity

t = time.

If the ball is thrown from the ground then:

= 0 m

We want to find now the distance between the window and the ground.

When the ball reaches the bottom of the window t = 1.8s

So:

m

The window is 37,404 m high

Finally, to know how high the ball rises we must know at what moment the vertical velocity of the ball is zero.

Now we replace t in the position equation

The ball reaches up to 44.70 m in height.

double the distance = divide strength by 4