The hieght in feet h of an object thrown upwards... HELP ME?

Question

This is my last question for my homework! help me!

The hieght in feet H of an object thrown upwards from the ground as a function of the time t in seconds since being released is given by

h=-16t+128t

Determine the time in seconds required for the object to reach a hieght of 40 feet on it's way up?

Okay, basically, I don't have a clue what this is asking. If you can explain how you got your answer, i'll give you the best answer

Answer 1

My question to you is are you sure that one of the t's is not supposed to be squared? I'm assuming that your problem is in feet, leaving the acceleration at -32 ft/s2. If that is the case than this object is thrown up at 128 ft/sec. t=0.33 seconds. on the way up , and t = 7.7 on the way down.

Answer 2

*You typed the equation wrong. Its > h = -16t^2 + 128t in which, the 1st "t-variable" is squared. You have to find the time when the object reaches 40 feet.

First: replace "40" with the "h" variable....

40 = -16t^2 + 128t

*Set the equation to "0" > subtract 40 from both sides...

40 - 40 = -16t^2 + 128t - 40
0 = - 16t^2 + 128t - 40

Sec: Factor > the 1st coefficient is negative so, factor out a negative sign with the number that's divisible with the coefficients which is, - 8

0 = - 8(2t^2 - 16t + 5)

*Factor the expression in the parenthesis, since its not possible to factor: use the Quadratic Formula.

x = [- b +/- V`(b^2 - 4ac)] / 2a

*Take the coefficients and replace them with the corresponding variable: a = 2 > b = - 16 > c = 5

x = [- (-16) +/- V`((-16)^2 - 4(2)(5))] / 4

x = [16 +/- V`((-16)(-16) - 4(2)(5))] / 4

x = [16 +/- V`(256 - 4(2)(5))] / 4

x = [16 +/- V`(256 - 4(10))] / 4

x = [16 +/- V`(256 - 40)] / 4

x = [16 +/- V`(216)] / 4

x = [16 +/- V`(2*2*2*3*3*3)] / 4

x = [16 +/- 2(3)V`6] / 4

x = [16 +/- 6V`6] / 4

*You will have two solutions - one has addition, the other has subtraction.

1. x = [16 + 6V`6] / 4
x = (16/4) + (6/4)V`6
x = 4 + (3V`6)/2

2. x = [16 - 6V`(6)] / 4
x = (16/4) - (6/4)V`6
x = 4 - (3V`6)/2

Solutions: 4 + (3V`6)/2 and 4 - (3V`6)/2

Answer 3

Hi, when a mass is thrown vertically upwords with an initial speed vo in the earth gravitational field with g (m/s^2) as acceleration than the following equation describes the height h(m) of the mass as function of time t(s)

h = -1/2*g*t^2 + vo*t m

1. In your equation the factor in the first term -16 must be -1/2*g.

2. In the first term must be t^2

3. 128 must be the initial speed and it has a dimension,inyour case measured in f/s, which would be with 1ft = 0,3048m
128ft/s = 39 m/s.

40*0,3048 = 12,19 m

12,19 = 1/2*9,81*t^2 +39*t = h (in m)

Multplying everything leads us to a quadratic equation

12,19 = -4,9*t^2 + 39*t

Rearranging the terms gives us

t^2 - 39/4,9*t = -12,19/4,9 = t^2 - 7,96*t = -2,49

The two solutions are

t1/2 = 7,96/2 +/- sqrt((7,96/2)^2 -2,49))

t1 = 3,98 + sqrt((3,98^2-2,49)) = 7,63 s

t2 = 3,98 - sqrt((3,98^2 - 2,49)) = 0,33 s

We get two solutions, because a specific height is reached in the upward and in the downward motion.

Answer 4

h=-16t+128t
First correct the equation to be:
h=-16t^2+128t , where h is height and t is time in seconds.
40 = -16t^2 +128t, or
-16t^2 +128t -40= 0
Use quadratic formula:
t = [-128 +/- sqrt(128^2-4(-16)(-40)]/-32
t = [-128 +/- 117.6]/-32
t= [-128+117.6]/-32 =.325 seconds, or
t = (-128-117.6]/-32 = 7.675
So the object will reach a height of 40 feet in .325 seconds and then will reach a height of 40 feet again on its way back down after 7.675 second have alapsed

Answer 5

Just set h=40 and solve for t. h is in SI units as it is.

Assuming you meant h= -16t^2+128t and not h=-16t+128t...

40 = -16t^2 + 128t

16t^2-128t+40=0

4t^2 - 32t + 10 = 0

2t^2 - 16t + 5 = 0

t = [8+3sqrt6]/2 or t= [8-3sqrt6]/2

Answer 6

You sure that's the formula? No "t^2" instead of "t" for one of those terms? Because what you wrote just simplifies to h = 112t, meaning the height is just going to increase with t, and the object you're throwing never comes back down.

Answer 7

well first h=112t
your teacher wants you to work out how long it would take to get to a height of 40 so we need to work out 40 h rather than just 1h...
h=112t
40h=(40)112t
which is 4480
unless as the other man said you typed it wrong p.s. how old r u? plz add 2 ur q