Object Word Problem

Question

An object was shot up into the air at an initial vertical speed of 640 feet per second. Its height as time passes can be modeled by the quadratic function f, where f(t)=−16t^2+640t. Here t represents the number of seconds since the object’s release, and f(t) represents the object’s height in feet.

1) After blank, this object reached its maximum height of blank

 2) This object flew for blank before it landed on the ground.

3) This object was blank in the air 23 seconds after its release.

4) This object was 3264 ft high at two times: once blank after its release, and again later blank  after its release.

Answer 1

Since this question is not listed under Calculus, let's solve the max height question using the fact that the graph of this equation is a downward opening parabola (leading coefficient is negative), so reaches its maximum height at its vertex.

One way to locate the vertex is to use the fact that the axis of symmetry is given by x = -b/2a. In this case a = -16 and b = 640.

So the axis of symmetry is x = -640/-32 = 20. The vertex lies on the axis of symmetry, so the x coordinate of the vertex is 20. To find the y-coordinate (which will be the max height) find f(20).

F(20) = -16(20)² + 640(20) = 6400 feet.

After 20 seconds this object reached is maximum height of 640 feet.

The object will reach the ground when the height is 0.

-16t² + 640t = 0

-16t(t - 40) = 0

t = 0 or t = 40 (t = 0 is the initial time the object was at ground level)

Note: You could also say it took the object 20 seconds from ground level to reach its high point, so another 20 seconds to fall back to earth.

The object flew for 40 seconds before it landed on the ground.

To find how high the object was after 23 seconds, find f(23):

f(23) = -16(23)²+640(23) = 6256 feet

To determine the times when the object was 3264 feet in the air solve this equation:

-16t²+640t = 3264

16t²-640t+3264 = 0

Here I would check to see if 3264 is evenly divisible by 16 (which it is).

t² - 40t + 204 = 0

Are there two integers that multiply to give 204 and add to give - 40, i.e. does the left side factor?

If you jot down the prime factors of 204 this might become apparent: 2, 2, 3, 17. Looks like 34 times 6 is 204, so:

(t - 34)(t-6) = 0

t = 34 or t = 6

6 seconds after its release and 34 seconds after its release.

Check it out here:

desmos.com/calculator/dboduxfl1m

Answer 2

h= -16t^2 +640t

h' = -32t +640 = 0

t= 640/32 = 20 seconds to reach maximum height

it hit the ground in twice that time = 40 seconds

max height was h(20) = -16(20)^2 +640(20) = -6400+12800 = 6400 feet high

after 23 seconds it's 3 seconds after its max height, coming down, so it's h<6400 ft but not much less.

h(23) = -16(23)^2 + 640(23) = -8464 + 14720 = 6256 feet high after 23 seconds

3264 = -16t^2 +640t

16t^2 -640t +3264 = 0

divide by 16

t^2 -40t +204 = 0

factor, complete the square or use the quadratic formula

t^2 -40t + 400 = 400-204 =196

(t-20)^2 = 196

t = 20 +/-sqr196

t = 20+/-14

t = 20-14, 20+14

t = 6 or 34 seconds when it's 3246 feet high, going on the way up after 6 seconds and on the way down after 14 seconds