A rocket fired into the air is modeled by the function #h(t)=-16t^2 + 160t#. How many seconds is the rocket in the air?

2 Answers
Jun 8, 2016

The rocket will be in the air for #10# seconds.

Explanation:

We have to find out the positive values of #x# for which #h(x)>=0#.

We can do it algebraicly:

#-16t^2+160t>=0#

#-16t*(t-10)>=0#

#t_1=0, t_2=10#

Since the coefficient next to #x^2# is nagative (#-16#), the solution of this inequality is #t in <0;10>#, so the rocket's flight will last #10# seconds.

We can also solve this task using graph of the function:

graph{-16x^2+160x [-50, 50,-20, 428]}

From this graph we clearly see, that the height is not lower than zero for #t in <0;10>#

Jun 8, 2016

#10s#

Explanation:

Given is
#h(t)=-16t^2+160t#
It has not been explicitly given but it is presumed that #h# is the height of rocket and #t# is time in seconds.

Clearly the rocket will be in air between the time interval from #t_0=0# and #t=t_1# when height of the rocket is zero.
Setting the height #h(t)=0# and solving for #t#
#-16t^2+160t=0#
Factorizing we obtain
#-16t(t-10)=0#
We have two values of #t#

  1. from first factor
    #t=0#
  2. from second factor
    #(t-10)=0#
    #=> t=10#

We obtain #t_1-t_0=10# seconds is time for which the rocket is in the air.
Graphically
graph{y=-16x^2+160x [-5, 15,-15, 428]}
height #h# is plotted as #y# and time #t# as #x#
We obtain time of flight as #t=10#