need help tell me how i got my answers, i'm confused?

a projectile is fired upward from the ground with an initial velocity of 844ft.per/sec.Assume that the height of the projectile @ any time t can be describe by the given polynomial function.P(t)=-16t^2+844t. find the height of the projectile at each given time. 3 secs.I have ( 2388), 6 sec.(4488), 9sec.(6300) and 11 sec. (7348). explain why the height changes at time passes. ( as time increases,a poly. of degree 2 and neg. coefficient of t^2 increases,reaches a maxium and then decreases. object hit the ground a sec.( 53 sec.)

i need to get help on how to do this. I dont fully understand but to multiply each sec. by the t's. and then I'm lost.. please help

3 Answers

  • gp4rts
    Lv 7
    1 decade ago
    Favorite Answer

    You have the right answers for each value of t. You must have done that by replacing t in the poynomial with its value. For example, to solve for t = 3 you will put 3 in wherever t appears. -16*(3)^2+844*3; this will give your result. At t increases, t^2 also increases, but t^2 increases faster than t. The height increases with t as long as 16*t^2 is less than 844*t; the squared term will eventually catch up and from that point the height decreases. This point is given by 16*t^2 = 844t; because this is a symmetrcal situation, the fall time is the same, so the total time to reach ground is twice the value of t gotten by solving that equation for t.

    If you do this you will see that the time to reach the ground is not 53 sec.

    • Commenter avatarLogin to reply the answers
  • Anonymous
    1 decade ago

    you will not gain anything if you ask other people to think for you

    you must do your own home work

    this may be an elaborate way of telling you that i dont know ,

    .but that is for you to figure out

    • Commenter avatarLogin to reply the answers
  • 1 decade ago

    I'm sorry, but i don't know. but that is really confused and it hurts my brain. i cant follow it, either.

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.