Question

A coin rolls off the edge of a table. The coin was traveling with a speed of

0.40 m/s. It lands 0.20 m away from the edge of the table. How high is the table?
(Hint: You must solve for time before you can solve for d)

Answer 1

Approximately
`1.3\; {\rm m}` (assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}`, the table is level, and that air resistance on the coin is negligible.)

Step-by-step explanation:

Assume that the air resistance on the coin is negligible. As long as the coin is in the air, its horizontal velocity of the coin will be constantly equal to the original value,
`v_(x) = 0.40\; {\rm m\cdot s^(-1)}`.

It is given that the horizontal displacement of this coin is
`x_(x) = 0.20\; {\rm m}` during the flight. Divide this horizontal displacement by horizontal velocity to find the duration
`t` of the flight:

`\begin{aligned} t &= (x_(x))/(v_(x)) \\ &= \frac{0.20\; {\rm m}}{0.40\; {\rm m\cdot s^(-1)}} \\ &= 0.50\; {\rm s}\end{aligned}`.

Also because air resistance on the coin is negligible, the vertical acceleration of the coin will be constantly
`a_(y) = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}`. If the table is level, the initial vertical velocity of the coin
`u_(y)` will be
`0\; {\rm m\cdot s^(-1)}`. Apply the SUVAT equation
`x = (1/2)\, a\, t^(2) + u\, t` to find the vertical displacement of the coin during the
`t = 0.5\; {\rm s}` that it is in the air:

`\begin{aligned}x_(y) &= (1)/(2)\, a_(y)\, t^(2) + u_(y)\, t \\ &= (1)/(2) \, (-9.81\; {\rm m\cdot s^(-2)})\, (0.5\; {\rm s})^(2) + (0\; {\rm m\cdot s^(-1)})\, (0.50\; {\rm s}) \\ &\approx (-1.3)\; {\rm m}\end{aligned}`.

In other words, the coin lands approximately
`1.3\; {\rm m}` below the table edge. Hence, the height of the table would be approximately
`1.3\; {\rm m}\!`.