# A rocket is launched from a platform that is 50 feet tall with an initial velocity of 120 feet per second. How high is the rocket after 2 seconds?

Jul 18, 2016

$225.6 f t$ above ground level

#### Explanation:

Treat it like two vectors. One up and the other down

The down is due to gravity and it has a downward acceleration of
32.2feet per second per second written as $32.2 \textcolor{w h i t e}{.} \frac{f t}{{s}^{2}}$

Let time count be $t$ seconds
Let distance upwards be ${d}_{u}$
Let distance downward be ${s}_{d}$
Let upward distance be positive
$\textcolor{m a \ge n t a}{\text{Let downward distance be negative}}$

'.......................................................
$\textcolor{b r o w n}{\text{Quick note about handling the units of measurement}}$

Any object that is in free fall with no other forces acting on it for a given time would have the velocity of:
$32.2 \times t \textcolor{w h i t e}{.} \frac{f t}{s} ^ \left(\cancel{2}\right) \times \cancel{s} \text{ " =" } 32.2 \times t \textcolor{w h i t e}{.} \frac{f t}{s}$ ......................Equation (1)
,........................................................................................
$\textcolor{b l u e}{\text{Upward distance if no gravity}}$

Upward velocity is constant so distance of upward vector is

$\implies {d}_{u} = 50 + 120 \times t \text{ "=" } 50 + 120 \times 2$

=>color(blue)(d_u=+290 feet .............................Equation (2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Downward distance due to gravity}}$

This is the mean velocity multiplied by time

Mean velocity $\frac{\text{Initial velocity + final velocity}}{2}$

Initial velocity is $0$

Final velocity [using Equation(1)]$\to 32.2 \times t = 32.2 \times 2 = 64.4 \frac{f t}{s}$

So mean velocity is $\frac{0 + 64.4}{2} = 32.2 \frac{f t}{s}$

$\implies \textcolor{b l u e}{{d}_{d} = 32.2 \frac{f t}{s} \times t = 32.2 \frac{f t}{\cancel{s}} \times 2 \cancel{s} = - 64.4}$
$\textcolor{b r o w n}{\text{This has a negative answer because down is negative}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Adding the distances}}$

$+ 290 - 64.4 = 225.6 f t$ above ground level