Question

The observation deck of tall skyscrapers 350m above the street. Determine the time required for a penny to free fall from the deck to the street below. Remember gravity has an acceleration of 9.8/s2 ______ seconds

Answer 1

Approximately
`8.5\; {\rm s}`, assuming that air resistance is negligible.

Step-by-step explanation:

Under the assumptions, acceleration of the penny during the flight would be constantly
`a = (-g) = (-9.8)\; {\rm m\cdot s^(-2)}` (negative because acceleration points downward towards the ground.) Additionally, the question implied that the initial velocity of the penny would be
`u = 0\; {\rm m\cdot s^(-1)}` by stating the penny was in a "free fall". The displacement of the penny during the flight is
`x = (-350)\; {\rm m}` (negative because the penny would be below the initial position.)

The duration of the flight can be solved in the following steps:

• Find velocity of the penny right before landing.
• Divide the change in velocity by acceleration to find the duration of the motion.

Since acceleration of the penny is constant, apply the following SUVAT equation to find the velocity right before landing:

`\displaystyle v^(2) - u^(2) = 2\, a\, x`,

Where:

• 
  `v` is the velocity right before landing,
• 
  `u = 0\; {\rm m\cdot s^(-1)}` is the velocity at the beginning of the motion,
• 
  `a = (-9.8)\; {\rm m\cdot s^(-2)}` is the acceleration of the penny, and
• 
  `x = (-350)\; {\rm m}` is the displacement of the penny.

Rearrange this equation to find
`v`:

`\begin{aligned}v &= \left(-\sqrt{u^(2) + 2\, a\, x}\right)\end{aligned}`.

(Note that the sign of
`v` should be negative because the coin would be travelling downward.)

The change in velocity would be
`(v - u)`. Divide this change in velocity by acceleration (the rate of change in velocity) to find the duration of the motion:

`\begin{aligned}t &= (v - u)/(a) \\ &= \frac{\sqrt{u^(2) + 2\, a\, x} - u}{a} \\ &= \frac{\left(-\sqrt{0^(2) + 2\, (-9.8)\, (-350)}\right) - 0}{(-9.8)}\; {\rm s} \\ &\approx 8.5\; {\rm s}\end{aligned}`.