Analysing the question:
We are given that the cat chases a rat on a 1m high table, this means that the height of the cat before falling is 1 m
The horizontal velocity of the cat after sliding off the table will not change in mid-air since there is no force to oppose it
We are given:
height of the cat before falling (h) = 1 m
Distance covered by cat after falling off the edge = 2.3 m
acceleration due to gravity (a) = 9.81 m/s²
initial vertical velocity of the cat (u) = 0 m/s
Since the cat will fall at a constant horizontal velocity, the time taken by the cat to reach the ground times the constant speed of the cat will be equal to the horizontal distance covered by the cat
[ since time taken to reach the ground = time for which the constant horizontal velocity will be applied]
The above can simply be written as:
time taken by the cat to reach the ground * horizontal velocity = horizontal distance covered
Time taken by the cat to reach the ground:
From the second equation of motion
s = ut + 1/2*at²
replacing the variables with vertical components
1 = (0)(t) + 1/2 * (9.8)(t²)
1 = 4.9t²
t² = 1 / 4.9
t = 1 / √4.9
t = 0.45 seconds
Cat's horizontal velocity when it slid off the table:
Speed of cat = distance travelled / time taken
here, the distance is the horizontal distance covered by the cat after falling off the table
replacing the variables
Speed of the cat = 2.3 / 0.45 = 5.11 m/s
