# A ball thrown horizontally from a point 24 m above the ground, strikes the ground after traveling horizontally a distance of 18 m. With what speed was it thrown?

Nov 2, 2015

$8.14 \text{m/s}$

#### Explanation:

First of all we get the time of flight from the vertical component:

$s = \frac{1}{2} \text{g} {t}^{2}$

$t = \sqrt{\frac{2 s}{g}}$

$t = \sqrt{\frac{2 \times 24}{9.8}}$

$t = 2.21 \text{s}$

The horizontal component of velocity is constant so:

$v = \frac{s}{t} = \frac{18}{2.21} = 8.14 \text{m/s}$

Nov 2, 2015

see equation [5] and use your calculator :D

#### Explanation:

by setting the positive y-axis pointing upward and the positive x-axis pointing to the right,

[1] $y = {y}_{0} + {v}_{0 , y} \cdot t - 0.5 \cdot g \cdot {t}^{2}$
with ${y}_{0} = 24 m$

[2] $x = {x}_{0} + {v}_{0 , x} \cdot t$
with ${x}_{0} = 0$,

where
${v}_{0 , y} = 0$ and
${v}_{0 , x} = {v}_{0}$

at where the ball strikes,
$18 m = x = {v}_{0} \cdot t$ then, $18 = {v}_{0} \cdot t$
we then isolate t,
$t = \frac{18 m}{{v}_{0}}$

and insert t into (1) at the striking point, assuming that y = 0
[3] $0 = 24 + - 0.5 \cdot g \cdot {\left(\frac{18 m}{{v}_{0}}\right)}^{2}$
isolating ${v}_{0}$,
[4]${\left({v}_{0}\right)}^{2} = \frac{0.5 \cdot {\left(18 m\right)}^{2} \cdot g}{24}$
[5]${v}_{0} = \sqrt{\frac{0.5 \cdot {\left(18 m\right)}^{2} \cdot g}{24}}$

and the value of g depends on what you/your teacher uses ranging from [9,10].