# A stone is dropped from the peak of a hill. It covers a distance of 30m in last second of its motion. What is height of the peak?

Apr 15, 2017

$\text{peak height : "62.16" } m e t e r s$

#### Explanation:

$\text{let the arrival time of Object from A to B be t. }$

$A C = \frac{1}{2} g {t}^{2}$

$A B = \frac{1}{2} g {\left(t - 1\right)}^{2}$

$A C - A B = 30 \text{ } m e t e r s$

$30 = \frac{1}{2} g {t}^{2} - \frac{1}{2} g {\left(t - 1\right)}^{2}$

$30 = \frac{1}{2} g \left[{t}^{2} - {\left(t - 1\right)}^{2}\right]$

$\frac{30 \cdot 2}{g} = {t}^{2} - \left({t}^{2} - 2 t + 1\right)$

$\frac{60}{g} = \cancel{{t}^{2}} - \cancel{{t}^{2}} + 2 t - 1$

$\frac{60}{g} + 1 = 2 t$

$\frac{60 + g}{g} = 2 t$

$2 t = 7.12$

$t = \frac{7.12}{2}$

$t = 3.56 \sec$

$\text{peak height=} A C = \frac{1}{2} \cdot g \cdot {t}^{2}$

$A C = \frac{1}{2} \cdot 9.81 \cdot {\left(3.56\right)}^{2}$

$A C = 62.16 \text{ } m e t e r s$