Question

An object falls a distance h from rest. If it travels .50h in the last 1.00s, find (a) the time and (b) the height of its fall?

Answer 1

Height: `"57.1 m"`
Time: `"3.41 s"`

Explanation:

So, you know that your objects starts falling from rest from a height `h`. Moreover, you know that it covered a distance of `0.5h` in the last `"1.00 s"` of its motion.

Since this distance of `0.5h` was covered 1.00 seconds before reaching the ground, you can say that the distance it covered in the first part of its motion, from top position to `0.5h` is

`h_"first" = h - 0.5h = 0.5h`

So, it covered bottom half of its motion in one second, which means that you can write

`0.5h = v * "1 s" + 1/2 * g * "1 s"^2" "`, where

`v` - the velocity of the object after it travelled the first half of its motion.

`0.5h = v * 1 + 1/2 * 9.8 * 1 = v + 4.9`

Now focus on the first half of its motion. You can find a second relationship between `v` and `h` by writing

`v^2 = underbrace(v_0^2)_(color(blue)(=0)) + 2 * g * h_"first"`

`v^2 = 2 * g * 0.5h = g * h = 9.8 * h`

So, you have

`{(0.5h = v + 4.9), (v^2 = 9.8 * h):}`

Use the first equation to find `h` as a function of `v`, then plug that into the second equation to find `h`

`h = (v + 4.9)/0.5 = 2v + 9.8`

`v^2 - 9.8 * (2v + 9.8) = 0`

`v^2 - 19.6v - 96.04 = 0`

Use the quadratic formula to find the two solutions to this quadratic equation

`v_(1,2) = (-(-19.6) +- sqrt((-19.6)^2 - 4 * 1 * 96.04))/(2 * 1)`

`v_(1,2) = (19.6 +- sqrt(768.32))/2`

The negative solution has no physical significance in this context, which means that `v` will be

`v = (19.6 + 27.72)/2 = "23.66 m/s"`

Use this value to find `h`

`h = 2 * 23.66 + 9.8 = color(green)("57.1 m")`

To find the total time of flight, use

`h = underbrace(v_0)_(color(blue)(=0)) + 1/2 * g * t_"total"^2`

`t_"total" = sqrt((2 * h)/g) = sqrt((2 * 57.12color(red)(cancel(color(black)("m"))))/(9.8color(red)(cancel(color(black)("m")))/"s"^2)) = color(green)("3.41 s")`