Question

1. a cat chases a mouse across a 2.5 m - high platform. the mouse steps out of the way, and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. when the cat slid off the table, what was its speed?

Explanation:

Step1: Analyze vertical - motion

The cat's vertical - motion is a free - fall motion. The initial vertical velocity $v_{0y}=0\ m/s$, the acceleration due to gravity $g = 9.8\ m/s^{2}$, and the vertical displacement $y=- 2.5\ m$ (taking downwards as negative). Use the equation $y = v_{0y}t+\frac{1}{2}at^{2}$. Since $v_{0y} = 0\ m/s$, the equation simplifies to $y=\frac{1}{2}at^{2}$.
$y=\frac{1}{2}(-g)t^{2}$
$-2.5=\frac{1}{2}\times(-9.8)t^{2}$

Step2: Solve for time $t$

First, rewrite the equation from Step 1 for $t$:
$t^{2}=\frac{2\times2.5}{9.8}$
$t=\sqrt{\frac{2\times2.5}{9.8}}\ s$

Step3: Analyze horizontal - motion

The cat's horizontal - motion is a uniform - motion with constant velocity because there is no horizontal acceleration ($a_x = 0$). The horizontal displacement $x = 2.2\ m$. The horizontal velocity $v_x$ is constant, and we use the equation $x = v_x t$. So $v_x=\frac{x}{t}$.
Substitute $t=\sqrt{\frac{2\times2.5}{9.8}}$ into $v_x=\frac{x}{t}$:
$v_x=x\sqrt{\frac{g}{2y}}$
$v_x = 2.2\sqrt{\frac{9.8}{2\times2.5}}\ m/s$
$v_x=2.2\sqrt{1.96}\ m/s$
$v_x=2.2\times1.4\ m/s$
$v_x = 3.08\ m/s$

Answer:

$3.08\ m/s$