Question

Question #cd938

Answer 1

`1.53` `"s"`

Explanation:

We're (essentially) asked to find the time when the rock hits the ground, given it fell from a height of `11.5` `"m"`.

This rock is undergoing free-fall; i.e. it is under the sole influence of Earth's gravitational force, and falling toward Earth's surface with an acceleration of `-g`, which is `-9.8"m/s"^2`

To find the time `t` when the rock reaches a position of `y = -11.5` `"m"`, we can use the equation

`y = y_0 + v_(0y)t - 1/2g t^2`

where

• `y` is the position at time `t` (`-11.5` `"m"`),

• `y_0` is the initial position (`0` `"m"`),

• `v_(0y)` is the initial `y`-velocity, which is `0` since the rock was merely dropped,

• `t` is the time, in `"s"` (what we must find), and

• `g` is `9.8` `"m/s"^2` (the minus sign in front of the `1/2` indicates that this acceleration is downward)

Plugging in known values, we have

`-11.5` `"m" = 0` `"m" + (0` `"m/s)"t - (4.9"m/s"^2)t^2`

`(4.9"m/s"^2)t^2 = 11.5` `"m"`

`t^2 = (11.5"m")/(4.9"m/s"^2)`

`t = sqrt((11.5cancel("m"))/(4.9cancel("m")"/s"^2)) = color(red)(1.53` `color(red)("s"`

When dropped from a well with a depth of `11.5` meters, the rock will thus take `1.53` seconds to reach the bottom of the well.