# 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?

##### 1 Answer

Height:

Time:

#### Explanation:

So, you know that your objects starts falling from rest from a height

Since this distance of **before reaching the ground**, you can say that the distance it covered in the **first part of its motion**, from top position to

#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

**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^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 = (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 = (19.6 + 27.72)/2 = "23.66 m/s"#

Use this value to find

#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")#