# Question 94a89

Nov 28, 2017

OK, what we are facing is a kinematic question.

#### Explanation:

First job, list what you know in a neat table with units, put a question mark next to the thing you’re looking for and a dash next to the factors that are not mentioned:

s = ?m (displacement)
u = 15.9m/s (initial velocity)
v = 0m/s (final velocity, it stops at the highest point)
$a = - 9.81 \frac{m}{s} ^ 2$
(note the minus sign? It shows the acceleration vector is pointing in the opposite direction to the positive velocity vector)
t = -s

So we need one of the kinematic equations linking s, u, v and a.

There is only one: ${v}^{2} = {u}^{2} + 2 a . s$

as $v = 0$, ${v}^{2} = 0$

Rearranging gives: s = -u^2/(2a#

$s = - {15.9}^{2} / \left(2 \times - 9.81\right) = 12.9 m$ (to 3 s.f)

Nov 28, 2017

Use formulas $d = {v}_{0} t + \frac{1}{2} a {t}^{2}$ and ${v}_{f} = {v}_{0} + a t$

#### Explanation:

A key understanding here is that the final velocity in the y-direction will be zero.

${v}_{0}$=15.9 m/s and ${v}_{f}$= 0 m/s

First, calculate the time the stone will be in the air ${v}_{f} = {v}_{0} + a t$

$0 = 15.9 \frac{m}{s} + - 9.81 \frac{m}{s} ^ 2 \cdot t$

$- 15.9 \frac{m}{s} = - 9.81 \frac{m}{s} ^ 2 t$

$t = \frac{- 15.9 \frac{m}{s}}{-} 9.81 \frac{m}{s} ^ 2 = 1.62 s$

Then, $d = {v}_{0} t + \frac{1}{2} a {t}^{2}$

$d = 15.9 \frac{m}{s} \left(1.62 s\right) 12 + \frac{1}{2} \left(- 9.81 \frac{m}{s} ^ 2\right) 1.62 {s}^{2}$

d = 12.88 m