Algebra 2
- Thread starter stevieh
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mmm4444bot
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stevieh said:
The exercise clearly states the downward velocity. Do you understand the concept of velocity ?
I mean, I'm not sure why you're stuck, on the posted question. :?
BigGlenntheHeavy
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\(\displaystyle In \ general, \ the \ position \ of \ a \ free \ falling \ object \ (neglecting \ air \ resistance) \ under \ the\)
\(\displaystyle influence \ of \ gravity \ is:\)
\(\displaystyle s(t) \ = \ -16t^2+v_ot+s_o, \ v_o \ = \ initial \ velocity \ and \ s_o \ = \ initial \ height.\)
\(\displaystyle The \ above \ is \ known \ as \ the \ position \ function.\)
\(\displaystyle Now, \ in \ this \ problem \ we \ have: \ s(t) \ = \ -16t^2 \ +v_ot, \ s_o \ = \ 0.\)
\(\displaystyle Now \ the \ velocity \ function \ is \ just \ the \ derivative \ of \ the \ position \ function, \ hence:\)
\(\displaystyle v(t) \ = \ -32t+v_o.\)
\(\displaystyle Now \ in \ 2.75 \ seconds, \ the \ rocket \ reaches \ its \ maximum \ height \ of \ 550ft. \ and \ then \ falls \ back\)
\(\displaystyle to \ earth.\)
\(\displaystyle Also \ at \ 2.75 \ seconds, \ the \ velocity \ of \ said \ rocket \ is \ zero \ and \ then \ speeds \ up \ due \ to\)
\(\displaystyle pull \ of \ gravity.\)
\(\displaystyle Ergo, \ v(2.75) \ = \ 0 \ = \ -32(2.75)+v_o, \ \implies \ v_o \ = \ 88 \ ft./sec.\)
\(\displaystyle Therefore \ s(t) \ = \ -16t^2+88t\)
\(\displaystyle Now, \ when \ the \ rocket \ hits \ the \ ground, \ its \ position \ function \ = \ 0, \ ergo:\)
\(\displaystyle 0 \ = \ -16t^2+88t, \ \implies \ t \ = \ \frac{11}{2} \ seconds.\)
\(\displaystyle Hence \ v\bigg(\frac{11}{2}\bigg) \ = \ -32(11/2)+88 \ = \ -88 \ ft./sec \ (neglecting \ air \ resistance, \ the \ parachute).\)
\(\displaystyle Hence, \ if \ the \ rocket \ didn't \ have \ the \ parachute \ to \ slow \ down \ its \ velocity, \ it \ would\)
\(\displaystyle smash \ into \ the \ earth \ at \ a \ speed \ of \ 88 \ ft./sec.\)
\(\displaystyle Factoid: \ Speaking \ of \ rockets, \ what \ did \ the \ "V" \ stand \ for \ in \ Germany's \ VI \ and \ VII \ rockets?\)
\(\displaystyle influence \ of \ gravity \ is:\)
\(\displaystyle s(t) \ = \ -16t^2+v_ot+s_o, \ v_o \ = \ initial \ velocity \ and \ s_o \ = \ initial \ height.\)
\(\displaystyle The \ above \ is \ known \ as \ the \ position \ function.\)
\(\displaystyle Now, \ in \ this \ problem \ we \ have: \ s(t) \ = \ -16t^2 \ +v_ot, \ s_o \ = \ 0.\)
\(\displaystyle Now \ the \ velocity \ function \ is \ just \ the \ derivative \ of \ the \ position \ function, \ hence:\)
\(\displaystyle v(t) \ = \ -32t+v_o.\)
\(\displaystyle Now \ in \ 2.75 \ seconds, \ the \ rocket \ reaches \ its \ maximum \ height \ of \ 550ft. \ and \ then \ falls \ back\)
\(\displaystyle to \ earth.\)
\(\displaystyle Also \ at \ 2.75 \ seconds, \ the \ velocity \ of \ said \ rocket \ is \ zero \ and \ then \ speeds \ up \ due \ to\)
\(\displaystyle pull \ of \ gravity.\)
\(\displaystyle Ergo, \ v(2.75) \ = \ 0 \ = \ -32(2.75)+v_o, \ \implies \ v_o \ = \ 88 \ ft./sec.\)
\(\displaystyle Therefore \ s(t) \ = \ -16t^2+88t\)
\(\displaystyle Now, \ when \ the \ rocket \ hits \ the \ ground, \ its \ position \ function \ = \ 0, \ ergo:\)
\(\displaystyle 0 \ = \ -16t^2+88t, \ \implies \ t \ = \ \frac{11}{2} \ seconds.\)
\(\displaystyle Hence \ v\bigg(\frac{11}{2}\bigg) \ = \ -32(11/2)+88 \ = \ -88 \ ft./sec \ (neglecting \ air \ resistance, \ the \ parachute).\)
\(\displaystyle Hence, \ if \ the \ rocket \ didn't \ have \ the \ parachute \ to \ slow \ down \ its \ velocity, \ it \ would\)
\(\displaystyle smash \ into \ the \ earth \ at \ a \ speed \ of \ 88 \ ft./sec.\)
\(\displaystyle Factoid: \ Speaking \ of \ rockets, \ what \ did \ the \ "V" \ stand \ for \ in \ Germany's \ VI \ and \ VII \ rockets?\)