Question

A Cat is on a balcony floor (90cm below the railing), keenly eyeing a butterfly hovering 60 cm above the railing. With what speed must the cat leave the floor in order to arrive at the butterfly with the optimum cat pouncing speed of 0.45m/s?

Answer 1

We will have the following:

First, the equation to use is the following:

$d=v_ot+\frac{1}{2}at^2$

Now, we transform the total distance the cat would need to travel:

$90\operatorname{cm}+60\operatorname{cm}=150\operatorname{cm}\cdot\frac{1m}{100\operatorname{cm}}=1.5m$

So, the cat would need to travel 1.5 meters. ("d" in the equation).

Now, using the speed given we determine the time it would take the cat to traverse the 1.5 meters:

$t=\frac{1.5m\cdot1s}{0.45m}\Rightarrow t=\frac{10}{3}\Rightarrow t=3.333\ldots$

So, the time it would take the cat to traverse the distance will be approximately 3.33 seconds.

Now, we know that the acceleration will be given by Earth's gravity, so:

$1.5m=v_0(\frac{10}{3}s)+\frac{1}{2}(-\frac{9.8m}{s^2})(\frac{10}{3}s)^2\Rightarrow1.5m=v_0(\frac{10}{3}s)+(-\frac{490}{9}m)$$\Rightarrow\frac{1007}{18}m=v_0(\frac{10}{3}s)\Rightarrow v_0=\frac{1007}{60}\frac{m}{s}\Rightarrow v_0=16.78333\ldots$

So, the initial vvelocity the cat must leave the floor in order to arrive at the butterfly with the optimum pouncing speed of 0.45 m/s is approximately 16.78 m/s or exactly 1007/60 m/s.