Question

Question #79d47

Answer 1

`"21.3 m"`

Explanation:

The first thing to do here is to figure out the time needed for the rock to hit the ground.

Your tool of choice here is the equation

`color(blue)(ul(color(black)(h = v_0 * t + 1/2 * g * t^2)))`

Here

• `h` is the distance covered by the object in a given time `t`
• `v_0` is the initial velocity of the rock
• `t` is the time of flight
• `g` is the gravitational acceleration, usually given as `"9.81 m s"^(-2)`

In your case, the object starts from rest, so

`v_0 = "0 m s"^(-1)`

This means that you will have

`h = 1/2 * g * t^2`

Rearrange to solve for `t`

`t = sqrt((2 * h)/g)`

Plug in your values to find

`t = sqrt( (2 * 28.5 color(red)(cancel(color(black)("m"))))/(9.81color(red)(cancel(color(black)("m")))"s"^(-2))) = "2.41 s"`

So, you know that it takes `"2.41 s"` for the rock to hit the ground. Use this value to determine the time spent falling before reaching the `"1.2-s"` mark

`t_"needed" = "2.41 s" - "1.2 s" = "1.21 s"`

Use this value in the above equation and solve for `h`, the distance covered by the rock in the first `"1.21 s"` of falling

`h_"1.21 s" = 1/2 * 9.81 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-2)))) * (1.21)^2 color(red)(cancel(color(black)("s"^2)))`

`h_"1.21 s" = "7.18 m"`

Therefore, you can say that `"1.21 s"` after it began to fall and `"1.2 s"` before hitting the ground, the rock was

`color(darkgreen)(ul(color(black)(h_ "above ground" = "28.5 m" - "7.18 m" = "21.3 m"))) ->` above the ground