suppose a bullet is fired on a distant planet so that its height in feet after t seconds is given by?

Question

h=4t^2+16t+5
when is the bullet at its highest point?
when will the bullet strike the ground?
what is the max height the bullet will attain?

Answer 1

Edit: maybe the first term is negative. Gravity pulls downward.

To answer: (provided first term is negative)
(t should be greater than or equal to 0 to make sense)

when is the bullet at its highest point?
get the first derivative,set to zero solve for t.

when will the bullet strike the ground?
solve h=0, get the bigger t.

what is the max height the bullet will attain?
from the answer 1, substitute it back to h. that will be the height.

d:

Answer 2

I'm pretty sure you meant h = -4t^2 + 16t + 5, otherwise I guess it would be some weird anti-gravity planet! :D And I'm assuming the planet has no atmosphere because that's much easier. ^^

Anywho:

a) The bullet is at it's highest point when the derivative of h is equal to zero:

h'(t) = -8t + 16 = 0 Solve for t:

t = 2 So, the the bullet is at its highest point at 2 seconds.


b) The bullet will strike the ground when h = 0. So find the zeros of the function, which should be about: t = -0.29 & t = 4.29
Since -0.29 doesn't make sense for our problem, we'll ignore it and take t = 4.29.

The bullet will hit the ground after about 4.29 seconds.


c) How high will it go? Well, we know that the bullet is at its maximum height at t = 2, so plug 2 in for t and get your value:

h(2) = -4(2^2) + 16(2) + 5
= 21

So it will go 21 feet high.

There ya go! Be sure to check my answers because I'm an idiot and prone to mistakes! :D:D

Answer 3

note
the vertex (the highest point) is at t = -b/2a
t = -16/(2*4) = -2
to the bullet will be at the highest point in -2 seconds, which doesn't make too much sense does it? the equation should probably be
h = 4t^2 -16t + 5, to take into account the pull of gravity
which then makes the vertex
t = --16/(2*4) = 2 seconds

if at the vertex t = 2, then
h = 4*2^2 - 16*2 + 5 = 4*4 - 32 + 5 = -11 feet
which still does not make much sense, so there is still some problems with your equation. Seems like the planet aimed at is below the horizon........
how about if you try
h = -4t^2 + 16t + 5
h(2) = -4*2^2 + 16*2 + 5 = -4*4 + 32 + 5 = 21 feet as the max height

This equation does not factor so using quadratic formula

t = [-16 +/- sqrt(16^2 - 4(-4)(5) )]/(2(-4)
t = [-16 +/- sqrt(336)] /(-8)
t = [-16 + 18.33] / -8 = negative time, which is not rational so that leaves

t = [-16 - 18.33] / -8 = 4.29 seconds when it hits the ground

of course this is assuming h = -4t^2 + 16t + 5

Answer 4

The coefficient of t^2 is positive. Therfore this function has no maximum. It does have a minimum at t=2 of 37 but every other value of t produces a height that is > 37. The range is 37=
You probably left out the minus sign in front of the -4t^2 term.

Answer 5

the bullet at its highest point is on the angle it was fired. the bullet will strike the ground when theres no more propulsion to it and gravity is strong enough to pull it back. max height depends on gravity, and angle fired. since its a distant planet i dont care if a bullet fired there unless im going there to fire some bullets.

Answer 6

There is no negative acceleration in that equation so it will continue to get faster and faster (4t^2).

1) Never
2) Never
3) Infinity

(Try banging 3600 into your equation (1 hour) and see the result)