Question

A cat chases a mouse across a 1.2 m high table. The mouse steps out of the way and the cat slides off the table at a speed of 7.0 m /s. Where does the cat strike the floor (horizontal distance from table leg)?

Answer 1

Given: table’s height h = 1.2 m ; speed of cat $v = 7 \frac{m}{s}$; S = ?

when a cat chases a mouse and jumps off a table, its movement becomes the movement of a horizontally launched object. We find where does the cat strike (horizontal distance from table leg).

the cat movement like the picture below

the cat x and y movement equations

$S_y = -1.2m \space\space\space a_y = -g = 9.8\large\frac{m}{s^2} \space\space\space v_{oy} = 0$

$S = v_ot+\large\frac{at^2}{2}$

$S_y = v_{oy}t+\frac{1}{2}a_yt^2 \to H = \large\frac{gt^2}{2}$

from this equation we find $\space\space\space t =\sqrt{\large\frac{2H}{g}}$

$v_{ox} = 7\frac{m}{s} \space\space\space a_x = 0 \space\space \space\space\space\space S_x - ?$

we find $S_x$

$S_x = v_{ox}t+\frac{1}{2}a_xt^2 = v_{ox}t = v_o\sqrt{\large\frac{2H}{g}} = 7*\sqrt{\large\frac{2*1.2}{9.8}} = 1.2 m$

Answer: the cat strikes to the floor 1.2m distance from the table