Question 192340
This  is  a  physics  problem  under  "projectile  motion"

Sometimes  it  is  found  in  algebra  under  quadratic  topics  as  a  actual  application  example.

For  today,  we  are  asked  how  high  the  projectile  will  go,  in  a  given  time,  with  an  initial  velocity.    The  velocity  provides  the  upward  energy,  and  gravity  provides  the  downward  energy.
 
Usually,  the  angle  of  the  initial  velocity  is  important,  but  for  today  we  will  assume  straight  up.    

If  not,  let  me  know.

The  basic  eqn  is:

delta  y  =  v*t - (1/2) (32.2)  t^2

where : 
delta  y  is  change  in  height.  Note  as  time  progresses,  the  projectile  goes  up,  reaches  a  peak,  and  then  comes  down.  

v  is  initial  velocity.   units  in  this  case  are    ft per  sec or  ft/sec

t  is  elapsed  time.  units  are  sec

32.2  is  gravity.  we  are  using  English  today.  be  careful  of  metric.  english  units  are  32.2  ft  per  sec  squared,  or  32.2  ft/sec^2     (metric  is  9.8  m/sec^2)

1/2  is  a  constant  to  get  average  acceleration


1st  problem

y= vt-1/2 (32.2) t^2

    = (45) (3) -  1/2 (32.2) (3^2)

    =  135  - 144.9

    =  (-9.9)  ft

Note  we  started  at  10  ft  above  ground, therefore  10-9.9  =  (.1)  ft  above  ground

This  is  not  so  nice  a  first  problem.  The  projectile  has  been  in  the  air  long  enough  to  go  fully  up,  stop,  and  now  decend  to  just  above  ground   height.  

Additional  calc  shows  that  the  projectile  goes  up  to  a  peak  of  about  31.4  ft  in  1.4  sec  before  starting  down.

2nd  problem

y=vt-16.1t^2 =  95*2  -16.1 *  2^2=  190-64.4=125.6  ft

Again,  initial  height  was  50  ft,  therefore  final  position  is  50+125.6 =175.6   ft

additional  calc  shows  peak  at  about  3 sec  of  about  140 ft