Question

stone thrown vertically upward with speed of 15.5m/s from edge of cliff 75.0m high. how much later does it reach bottom of cliff?

Answer 1

Approximately
`5.66\; {\rm s}`, assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}` and that air resistance is negligible.

Step-by-step explanation:

Let upward be the positive direction. Under the assumptions, acceleration of the stone would be
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}` (negative since the stone is accelerating downward.)

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

• Find velocity right before landing given displacement, initial velocity, and acceleration.
• Find duration of the flight from acceleration and the change in velocity.

In SUVAT equation
`v^(2) - u^(2) = 2\, a\, x`:

• 
  `v` is the final velocity right before landing (needs to be found,)
• 
  `u = 15.5\; {\rm m\cdot s^(-1)}` is the initial velocity,
• 
  `a = (-9.81)\; {\rm m\cdot s^(-2)}` is acceleration, and
• 
  `x = (-75.0)\; {\rm m}` is displacement (downward because the stone landed below where it was launched.)

Rearrange this equation to find
`v`:

`\begin{aligned}v &= \sqrt{u^(2) + 2\, a\, x} \\ &= \sqrt{(15.5)^(2) + 2\, (-9.81)\, (-75.0)} \; {\rm m\cdot s^(-1)} \\ &\approx -40.05\; {\rm m\cdot s^(-1)}\end{aligned}`.

In other words, the velocity of this stone has changed from the initial value of
`u = (15.5)\; {\rm m\cdot s^(-1)}` to
`v \approx (-40.05)\; {\rm m\cdot s^(-1)}` during the flight. Divide the change in velocity by acceleration
`a = (-9.81)\; {\rm m\cdot s^(-2)}` (the rate of change in velocity) to find the duration of the flight:

`\begin{aligned}t &= (v - u)/(a) \\ &\approx ((-40.05) - (15.5))/((-9.81))\; {\rm s} \\ &\approx 5.66\; {\rm s}\end{aligned}`.

In other words, the stone would be in the air for approximately
`5.66\; {\rm s}`.