suppose a bullet is fired on a distant planet so that its height(in ft.)after t seconds is given h=-4t^2+16t+5

Question

(ALGEBRA-QUADRATIC MODELS):
A. when is the bullet at its highest point?
B.when will the bullet strike the ground?
C.what is the maximum height the bullet will attain?
thanks alot guys! :)

Answer 1

Calculus tells us a lot about optimizing functions. However, it isn't needed here. No matter what planet you're on, the height-vs-time graph of a free-falling object is always a parabola.

There are a lot of facinating facts about parabolas and conic sections in general, but let's get to answering the questions that you asked. ;)

The maximum or minimum of a parabola occurs at its vertex. The value of the independent variable ( t ) at the vertex of a parabola is -B/2A. Therefore, t = -16/2(-4) = -16/-8 = 2.

A) The bullet reaches its highest point two seconds after it is fired.

Moving on, the height of the bullet it determined by the formula itself: h = -4t^2 + 16t + 5. If the bullet is on the ground, then h = 0. So:

-4t^2 + 16t + 5 = 0
4t^2 - 16t - 5 = 0 [Multiply by -1 to work with leading positive.]
t = [16 +/- sqrt(16^2 - 4(4)(-5))]/(2*4) [Quadratic Formula]
t = [16 +/- sqrt(336)]/8

The negative root does no good, as the bullet cannot land before it was fired. But the positive root show us that:

B) The bullet hits the ground approximately 4.3 seconds after is was fired.

Last is actually the easiest part, although most students miss it. You know the bullet reaches its highest point when t = 2 from part A. What will be its height at that point? You already have the equation. h = -4t^2 + 16t + 5.

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

C) The bullet will reach a height of 21 feet.

So, to sum up:

A) The bullet reaches its highest point two seconds after it is fired.
B) The bullet hits the ground approximately 4.3 seconds after is was fired.
C) The bullet will reach a height of 21 feet.

I hope this helps! =]

Answer 2

Graph the function. h at the vertex is the maximum hieght. Or put the equation in the form:

h=-4t^2+16t+5

h = -4(t^2 - 4t) +5

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

h = -4(t-2)^2 + 21

h = -4(t-2)^2 + 21

c) Hint: Lets focus on the right side. Since the first term is negative all it can do is decrease the value of h. If we vary t such that the first term is zero, we have the maximum h.

b) The bullet will strike the ground if h=0, solve for t.

a) Hint: substitute the valu for maximum h and solve for t.


Or for a and c, the vertex of the parabola represent the max h and t at the max h.

Answer 3

A. h = -4t^2 + 16t + 5
=> dh/dt = -8t + 16
at max height dh/dt = 0
=> -8t + 16 = 0
=> t = 2
Therefore, the bullet is at its highest point after 2 sec.

B. at the ground h = 0
=> -4t^2 + 16t + 5 = 0
=> t = (16 ± √(256+80))/8 = (16 ± 18.33)/8 = 4.29
(the negative root is extrenuous as time cannot be negative)
Therefore the bullet strikes the ground after 4.29 sec of the firing.

C. maximum height = -4.2^2 + 16.2 + 5 = 21 feet

Answer 4

h=f(t)=-4t^2+16t+5
Diffrentiate
f'(t)=-8t+16 and f''(t)=-8 hence maxima
Equate f'(t)=0
=> t= 2
So at t= 2 seconds bullet is at highest point
and the maximum height is -4*4+16*2+5=21
solve f(t)=0 to get when bullet strikes the ground
t= (-16+-sqrt(256+80))/(-8)
=(-16+-18.33)/ (-8)

you have t=-0.29 and t=4.29125
So at 4.29 sec bullet strikes the ground

Answer 5

A). h must be maximal
h=-4t^2+16t+5
h = -4(t2 - 4t + ..) + 5
h = -4(t2 - 4t + 4) + 5 +16
h = 21 - 4(t-2)^2
At t=2 s is h maximal 21 ft.

B) Two s later it hits the ground.

C) See A)

Answer 6

Just don't fire the bullet straight up in the air coz it will fall back on your head with double the speed.

Answer 7

huh?? was dat in english??