consider the motion of the cat along the vertical direction after sliding off the table
Y = vertical displacement of the cat from top of the table to the ground = - 1.3 m
v₀ = initial velocity along the Y-direction of the cat = 0 m/s
a = acceleration of the cat = acceleration due to gravity = - 9.8 m/s²
t = time taken to hit the ground
using the kinematics equation
Y = v₀ t + (0.5) a t²
- 1.3 = 0 t + (0.5) (- 9.8) t²
- 1.3 = - 4.9 t²
t = 0.51 sec
consider the motion of the cat along the horizontal direction :
v = velocity along the horizontal direction =
t = time of travel = 0.51 sec
X = horizontal displacement of the cat = 0.75 m
using the kinematics equation
X = v t (Since there is no acceleration along the horizontal
direction)
inserting the values
0.75 = v (0.51)
v = 1.45 m/s