Question

A cat chases a mouse across a 0.57 m high

table. The mouse steps out of the way, and
the cat slides off the table and strikes the floor
2.8 m from the edge of the table.
The acceleration of gravity is 9.81 m/s².
What was the cat's speed when it slid off
the table?
Answer in units of m/s.

Answer 1

8.21 m/s

Step-by-step explanation:

First we have to determine the time it took for the cat to hit the floor

We will have to assume that the cat's motion was entirely in the horizontal direction when it slid off the table. This is not an unreasonable assumption

Therefore when the cat slid off the table, its vertical velocity was 0

We will use the formula

`h = ut + (1)/(2)at^2`
to find the time taken for the cat to hit the floor from a height of 0.57 m

where

h = height of the table = 0.57m

u = initial vertical velocity = 0

a = acceleration = gravitational acceleration = g = 9.8 m/s²

Plugging in values we get

`0.57 = 0\cdot t + (1)/(2)\cdot 9.8\cdot t^2`

⇒
`\rightarrow 0.57 = 4.9t^2\\\\\rightarrow t^2 = (0.57)/(4.9)\\\\\rightarrow t = √(0.1163265)\\\\\rightarrow t \approx 0.341067 \textrm{ s}\\\\\\`

The horizontal distance traveled by the cat is 2.8m. The time it takes to travel this distance must be the same as the time it took the cat to hit the floor. There is no acceleration in the horizontal direction

So the velocity in the horizontal direction

=
`\rm (distance)/(time)$`

`= (2.8)/(0.341067)`

≈ 8.209537 m/s ≈ 8.21 m/s

Answer 2

3.37

Step-by-step explanation: