Question

A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 2.9 m from the edge of the table. The acceleration of gravity is 9.81 m/s 2 . What was the cat’s speed when it slid off the table? Answer in units of m/s.

Answer 1

At the start, we know the range of the cat's jump (R') and the height of the cat when it was on the table (H'), by using equations for both displacements we can find the initial velocity of the cat or $v_{0_x}$ and then we find the following equations:

$R'=2.9\,m \\ H'=1\,m \\ H'=\frac{1}{2}gt^2_{\text{cat falling}} \\ \implies t_{\text{cat falling}}=\sqrt{\frac{2H'}{g}} \\ R'=v_{0_x}t_{\text{cat falling}} \\ \implies v_{0_x}=\frac{R'}{t_{\text{cat falling}}} ={R'}\sqrt{\frac{g}{2H'}}$

We substitute to find the initial speed of the cat when it slid off the table as:

$v_{0_x} ={R'}\sqrt{\frac{g}{2H'}}={(2.9\,m)}\sqrt{\frac{9.81\,m/s^2}{2(1\,m)}} \\ v_{0_x}=6.4227\,m/s$

In conclusion, the cat slid off the table at around $v_{0_x}=6.423\,m/s$ .

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.