Question

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?

Answer 1

`225.6 ft` 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 color(white)(.)(ft)/(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
`color(magenta)("Let downward distance be negative")`

'.......................................................
`color(brown)("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:
`32.2xx t color(white)(.) (ft)/s^(cancel(2))xx cancel(s)" " =" " 32.2xxtcolor(white)(.) (ft)/s` ......................Equation (1)
,........................................................................................
`color(blue)("Upward distance if no gravity")`

Upward velocity is constant so distance of upward vector is

`=>d_u=50 + 120 xxt " "=" "50+120xx2`

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

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

`color(blue)("Downward distance due to gravity")`

This is the mean velocity multiplied by time

Mean velocity `("Initial velocity + final velocity")/2`

Initial velocity is `0`

Final velocity [using Equation(1)]`-> 32.2xxt =32.2xx2 = 64.4 (ft)/s`

So mean velocity is `(0+64.4)/2= 32.2 (ft)/s`

`=>color(blue)(d_d=32.2 (ft)/sxxt = 32.2 (ft)/(cancel(s)) xx 2cancel(s)=-64.4)`
`color(brown)("This has a negative answer because down is negative")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Adding the distances")`

`+290-64.4 = 225.6 ft` above ground level