Question

7- a ball is thrown vertically downward with an initial velocity from the top of a 36.0 m - tall building. the ball passes the top of a window that is 12.0 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?

Explanation:

Step1: Determine the displacement

The ball starts from the top of a 36.0 - m tall building and passes the top of a window 12.0 m above the ground. So the displacement $y$ is $y=36 - 12=24$ m. The time $t = 2.00$ s and the acceleration due to gravity $g=9.8$ m/s². We use the kinematic equation $y=v_0t+\frac{1}{2}gt^{2}$, where $v_0$ is the initial velocity.
$y = v_0t+\frac{1}{2}gt^{2}$
$24=v_0\times2+\frac{1}{2}\times9.8\times2^{2}$
$24 = 2v_0+19.6$
$2v_0=24 - 19.6$
$2v_0 = 4.4$
$v_0 = 2.2$ m/s.

Step2: Use the kinematic equation for final velocity

We use the kinematic equation $v = v_0+gt$ to find the velocity $v$ when it passes the top of the window.
$v=v_0+gt$
Substitute $v_0 = 2.2$ m/s, $g = 9.8$ m/s² and $t = 2$ s.
$v=2.2+9.8\times2$
$v=2.2 + 19.6$
$v=21.8$ m/s.

Answer:

$21.8$ m/s