Question

Tom the cat is chasing Jerry the mouse across a table surface 1.5 m off the floor. Jerry steps out of the way at the last second, and Tom slides off the edge of the table at a speed of 5.0 m/s. Where will Tom strike the floor, and what speed will Tom have before hitting the floor

Answer 1

Tom the cat is chasing Jerry the mouse across a table surface $1.5\mathrm{m}$ off the floor. Jerry steps out of the way at the last second, and Tom slides off the edge of the table at a speed of $5.0~\mathrm{m / s}$. Where will Tom strike the floor, and what speed will Tom have before hitting the floor.

Solution

$\Delta Y$ is the change in the y direction which is $1.5\mathrm{m}$ in the problem. We know that the acceleration in the y direction is $9.81\mathrm{m / s}$. We also know that Tom goes off the table in the horizontal or x direction so his initial velocity in the y direction has to be 0.

Let's use the kinematics equation:

$$\Delta Y = V _ {i} t + \frac {1}{2} g t ^ {2} = 0 * t + \frac {1}{2} g t ^ {2} = \frac {1}{2} g t ^ {2}.$$ $$t = \sqrt {\frac {2 \Delta Y}{g}} = \sqrt {\frac {2 * 1 . 5}{9 . 8 1}} = 0. 5 5 s.$$

Now we need to solve for $\Delta X$, or the change in the $x$ direction. This will tell you how far from the table Tom lands. We use the formula:

$$\Delta X = V _ {0} t = 5. 0 \frac {\mathrm {m}}{\mathrm {s}} * 0. 5 5 \mathrm {s} = 2. 7 5 \mathrm {m}.$$

The speed of Tom before hitting the floor:

$$V _ {f} = \sqrt {V _ {x} ^ {2} + V _ {y} ^ {2}}.$$

We know that $V_{x} = V_{0} = 5.0\frac{\mathrm{m}}{\mathrm{s}}$ and $V_{y} = gt$. So

$$V _ {f} = \sqrt {V _ {0} ^ {2} + (g t) ^ {2}} = \sqrt {5 . 0 ^ {2} + 9 . 8 1 ^ {2} * 0 . 5 5 ^ {2}} = 7. 4 \frac {m}{s}.$$

Answer: 2.75 m; 7.4 $\frac{m}{s}$.