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
[tex]v = \frac{d}{t}[/tex]
[tex]v = \frac{2}{0.20}[/tex]
[tex]v = 10\ \frac{m}{s}[/tex]
We know that:
[tex]v = v_0 -gt[/tex]
Where [tex]v_0[/tex] is the initial velocity.
So:
[tex]10 = v_0 -9.8(2)[/tex]
[tex]10 + 9.8(2) = v_0[/tex]
[tex]v_0 = 29.6\ m / s[/tex]
Then we have the equation for the position as a function of time.
[tex]r = r_0 + v_0t - \frac{1}{2}gt^2[/tex]
Where
[tex]r_0[/tex] = initial position
r = position as a function of time
[tex]v_0[/tex] = initial velocity
g = acceleration of gravity
t = time.
If the ball is thrown from the ground then:
[tex]r_0[/tex] = 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:
[tex]r = 0 + 29.6(1.8) - 0.5(9.8)(1.8)^2[/tex]
[tex]r = 37,404[/tex] 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.
[tex]v = v_0 -gt\\\\0 = v_0 -gt\\\\gt = v_0\\\\t = \frac{29.6}{9.8}[/tex]
[tex]t = 3.02\ s[/tex]
Now we replace t in the position equation
[tex]r = 0 + 29.6(3.02) -0.5(9.8)(3.02)^2[/tex]
[tex]r = 44.70\ m[/tex]
The ball reaches up to 44.70 m in height.
