Question

Question #a307c

Answer 1

The total time of its fall is equal to `t_"total" = (2+sqrt(2))"s"`.
The total distance covered is equal to `h = 10(3 + 2sqrt(2))"m"`.

Explanation:

I think that you can go about solving this one in others ways as well, but here's the approach I used.

You know that the object covered half of its total path in 1 second. Moreover, it was the last second of its fall. This means that the distance it covered is actually the bottom half. You can write

`underbrace(h/2)_(color(blue)("second half of the fall")) = v_1t + 1/2g * t^2`, where

`h` - the total distance of the fall;
`v_1` - the speed with which it enters the bottom half of the total distance;
`t` - the time it took for this distance to be covered - in your case, `t` is equal to 1 second;
`g` - the gravitational acceleration, which I'll use as `"10 m s"^(-2)`

Use the value of `t` to determine a relationship between `h/2` and `v_1`.

`h/2 = v_1 * 1 + 1/2 * 10 * 1^2 = v_1 + 5` `" "color(blue)((1))`

Now focus on the first half of the fall. You know that your object covers the same distance, i.e. `h/2`, in a time I'll label as `t_1`.

Once again, you can write

`underbrace(h/2)_(color(blue)("first half of the fall")) = v_0 * t_1 + 1/2g * t_1^2`

Since the object starts from a stationary position, `v_0 = 0`. This will get you

`h/2 = 1/2 * 10 * t_1^2 = 5 * t_1^2` `" "color(blue)((2))`

For the first half of the fall, you can also write this equation

`v_1 = v_0 + g * t_1 <=> v_1 = g * t_1`

Use this identity to replace `v_1` in equation `color(blue)((1))` to get

`{ (h/2 = g * t_1 + 5 = 10t_1 + 5), (h/2 = 5 * t_1^2) :}`

Now solve for `t_1`.

`5t_1^2 = 10t_1 + 5 <=> t_1^2 - 2t_1 -1 = 0`

This quadratic has two solutions

`t_(1)^' = (2 + sqrt(4 + 4))/2 = 1 + sqrt(2)`

and

`t_(1)^('') = (2-sqrt(4 + 4))/2 = 1-sqrt(2)`

Since `t_1^('')` is negative, the only feasible solution is `t_1^(') = 1 + sqrt(2)`.

This means that the total time of the fall was

`t_"total" = t_1 + t = 1 + sqrt(2) + 1 = color(green)(2 + sqrt(2)" seconds")`

The total distance covered was

`h = 1/2 * g * t_"total"^2 = 5 * (2 + sqrt(2))^2`

`h = 5 * (4 + 4sqrt(2) + 2) = 30 + 20sqrt(2) = color(green)(10(3 + 2sqrt(2)" meters")`