Algebra II && Trig help please?

Question

The height, in feet, of a missile is given by h = 1600t - 16t^2 where t is the time in seconds.



A) What is the maximum height reached by the missile?
B) What is the time it reaches the maximum height?
C) What is the time when the missile hits the ground?

hey guys. I really need help with this math problem. I have 12 of these to do tonight, and I need one as an example. So could you please answer atleast one of the questions, and explain your steps so I can do the rest of the problems by myself? Thank you so much.

Answer 1

t=-b/(2a) is the vertex value so it's the time at which the max occurs

t=-1600/(2*-16)=50 sec

h(50) is the maximum height, just plug it in

Solve h=0 for the time when it hits the ground
1600t-16t^2=0
16t(100-t)=0
t=100 sec

Answer 2

If you know about function derive h= 1600t-16t^2

h'= 1600 -32t Put h'=0 1600 = 32t so t= 1600/32 =50s
(
This is the time it reaches maximum height (The 2nd derivative=
-32<0 so it is a maximum)

Put t=50 so h= 1600*50 -16*2500 =40,000 feet

If the missile is on the ground h must be zero so

1600t -16t^2=0 t(1600-16t)=0 which yields t= 0(starting point and t=1600/16=100s (time to reach the ground)

Answer 3

A) 40,000 feet
B) 50 secs
C) 100 secs


If you factor the equation 1600t - 16t^2, you get:
t x (1600 - 16t)

For h to equal 0, either the t has to be zero, or the (1600 - 16t) has to be 0

t = 0 the instant before you launch

For the other factor, 1600 - 16 t = 0,
or 1600 = 16t
1600 / 16 = 16t / 16
100 = t
so t = 100
This is the time when it hits the ground.

It is at max height when it is halfway through its parabolic (x^2 term) trajectory, or 50 secs.

Plugging in 50 secs you get:
h = (1600 x 50secs) - (16 x 50secs^2)
h = 40,000 feet


Good to see that you want an example so that you can finish the rest yourself, and learn the problem. Good luck.

Answer 4

I will help you...but I know you left this work undone until the very last minute!

h = 1600t-16t^2

a) the maximum height is determined by the maximum of the height equation. I will calculate this value using Calculus...you must plot the equation to find this same value.

h = 1600t - 16t^2

first derivative wrt time (dh/dt) = 1600 - 32t
The maximum occurs at the zero of this equation,

1600 = 32t

or t = 50 seconds (answer b)

at t = 50 seconds, the height is

h = 1600 (50) - 16(50)^2
h = 80000 - 40000 = 40000 ft

and because this is a parabolic flight, it takes the same amount of time to go up and then come down, so T = 50+50 or 100 sec.

You can also try solving this equation by completing the square...it will take too long. Just plot the equation and compute the values.

Good luck (you will need it if you have 11 of these babies to do!)

Answer 5

I think you have to answer B before you can answer A.

The equation for the height is a downward facing parabola (highest power on t is 2, and the coefficient of t is negative). The t that will maximize that equation is given by 1600 / (2*16) in this example. So the answer to part B is that the missile will reach it's max height when t=50 sec. If you plug 50 in for t in the equation for the height, you'll get 40,000 ft, the answer to A. To solve part C, set h=0 and solve for t.

1600t-16t^2=0
16t (100-t)=0
so t=0 or t=100.
You launched it at t=0, so it hits the ground at t=100 sec.

So, for a parabola y=ax^2 + bx +c, if a is positive the parabola opens up (has a minimum but no maximum). If a is negative, the parabola has a maximum but no minimum. The extreme point occurs in either case at x= -b/(2a).

Answer 6

A) The x coordinate of the vertex will be -b/2a=-1600/-32=50; the y coordinate of the vertex will occur at h(50)=1600*50-16*(50)^2=40000 feet @t=50 seconds. Finally, to answer the last question, note that h=t(1600-16t)=0 when t= 0 or t=100. The answer is t=100.

Answer 7

h(t) = 1600t - 16t^2
h(t) = -16t^2 + 1600t

t = (-b)/(2a)

t = (-1600)/(2(-16))
t = 100/2
t = 50

h(50) = -16(50)^2 + 1600(50)
h(50) = -16(2500) + 80000
h(50) = -40000 + 80000
h(50) = 40000

h = -16t^2 + 1600t
h = -16t(t - 100)

the intercepts are 0 and 100 so it takes 100 seconds to reach the ground.

ANS :
A.) 40000ft
B.) 50 seconds
C.) 100 seconds

Answer 8

Do you have a graphing calculator? If so, you might want to graph it on your graphing calculator.

After graphing, calculate the highest point on your parabola. The y-axis is the maximum height, which is 40,000 feet.

The y-axis is the time where the height is achieved, or 50 seconds.

The time the missile hits the ground is when the graph hits the x-axis, which is at (100,0), so the time when the missile hits the ground is 100 seconds.

Answer 9

carry on with the alg 2 and no trig till ur particularly a large math student which u reported it wasnt ur top-rated then confident. and confident trig in alg 2 wont be too undesirable yet whilst u get to precal or calc and u elect to take honors/ap for that then this is a killer type. and as quickly as u learn trig. it on no account leaves u.

Answer 10

first of all my gf's eigth grade kid isnt here right now to do this, but i will tell you that you need to give more info to solve this. You have to know at least two things, one of them is a value for t