# 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?

Sep 7, 2015

Height: $\text{57.1 m}$
Time: $\text{3.41 s}$

#### Explanation:

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

Since this distance of $0.5 h$ 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 $0.5 h$ is

${h}_{\text{first}} = h - 0.5 h = 0.5 h$

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

$0.5 h = v \cdot \text{1 s" + 1/2 * g * "1 s"^2" }$, where

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

$0.5 h = v \cdot 1 + \frac{1}{2} \cdot 9.8 \cdot 1 = v + 4.9$

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

${v}^{2} = {\underbrace{{v}_{0}^{2}}}_{\textcolor{b l u e}{= 0}} + 2 \cdot g \cdot {h}_{\text{first}}$

${v}^{2} = 2 \cdot g \cdot 0.5 h = g \cdot h = 9.8 \cdot h$

So, you have

$\left\{\begin{matrix}0.5 h = v + 4.9 \\ {v}^{2} = 9.8 \cdot h\end{matrix}\right.$

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

$h = \frac{v + 4.9}{0.5} = 2 v + 9.8$

${v}^{2} - 9.8 \cdot \left(2 v + 9.8\right) = 0$

${v}^{2} - 19.6 v - 96.04 = 0$

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

${v}_{1 , 2} = \frac{- \left(- 19.6\right) \pm \sqrt{{\left(- 19.6\right)}^{2} - 4 \cdot 1 \cdot 96.04}}{2 \cdot 1}$

${v}_{1 , 2} = \frac{19.6 \pm \sqrt{768.32}}{2}$

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

$v = \frac{19.6 + 27.72}{2} = \text{23.66 m/s}$

Use this value to find $h$

$h = 2 \cdot 23.66 + 9.8 = \textcolor{g r e e n}{\text{57.1 m}}$

To find the total time of flight, use

$h = {\underbrace{{v}_{0}}}_{\textcolor{b l u e}{= 0}} + \frac{1}{2} \cdot g \cdot {t}_{\text{total}}^{2}$

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")