Please help me solve this word problem (NEED HELP ASAP!!!!)

Question

The height of an object shot straight up with an initial velocity of v in feet per second is given by h + -16ft²+vt + c, where c is the initial height of the object above the ground. An object is shot up from 4 feet off the ground with an initial velocity of 48 feet per second. Will it reach a height of 40 feet? Use the discriminant to explain your answer.

Answer 1

c = 4 ft

v = 48 ft/s

 

We want to know if the height will be 40 at a certain time t.

 

h = 40

 

Plug in these values of c, v, and h into the equation.

 

40 = -16t² + 48t + 4

 

 

Subtract 40 on both sides of equation.

 

0 = -16t² + 48t - 36

 

 

We can factor out the 2 on the right side of the equation to simplify it.

 

h = 2(-8t² + 24t + 13)

 

 

The discriminant is the square-root part of the quadratic formula.  The value under the square-root is

 

b² - 4ac

 

where:

a = -8

b = 24

c = 13

 

24² - 4(-8)(13) =

576 + 416 =

992

 

The value under the square-root is positive.  Therefore, when we solve for t, we will have a real value of time.  In conclusion, the object will reach a height of 40 ft when we solve for that time.

 

 

 

Answer 2

This is a physics problem.

But if you want to solve it like it was a math problem - well, okay.

 

1st you can find the x-axis of symmetry.  x= -b/2a

Then you can find  f(-b/2a)

which will be the the y coordinate of the vertex

and so is the max height it will reach.

You need to know that 'c' is the beginning height and v is the initial velocity.

 

I don't think the discriminate tells you anything interesting in this problem.

It will tell you how many real or imaginary roots you have but that is not of interest in this problem.

 

btw - you have a (+) plus sign where there should be an = and other mistakes in the formula.

The formula is: h(t) = (1/2)(-32 ft/s²)(t²) +(v₀)(t) + h₀