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)?
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
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