Question

A classic physics problem states that if a projectile is

shot vertically up into the air with an initial velocity of
118 feet per second from an initial height of 80 feet off
the ground, then the height of the projectile, h, in feet, t
seconds after it's shot is given by the equation:
h 16t² + 118t + 80
=
Find the two points in time when the object is 112 feet
above the ground. Round your answers to the nearest
hundredth of a second.
The object is 112 feet off the ground at:
seconds.

Answer 1

The projectile reaches 112 feet at 8.28 seconds in the upward motion and in 7.69 seconds in downward motion

To find the two points in time when the object is 112 feet above the ground, we can set the height equation equal to 112 and solve for t:

ℎ =
`16t^(2)`+118t +80

h=112

Therefore,
`16t^(2)`+118t +80 = 112

`16t^(2)`+118t -32 = 0

Now, we have a quadratic equation in the form
`at^(2)`+bt+c=0, where

a=16,

b=118, and

c=−32.

We can solve for t using the quadratic formula:
`\frac{-b+-\sqrt{b^(2) - 4ac} }{2a}`

Therefore, t=
`\frac{-b+-\sqrt{b^(2) - 4ac} }{2a}`

Calculate the discriminant
`\sqrt{b^(2) - 4ac}}` =
`\sqrt{118^(2) - 4.16.(-32)}}` =
`√(15972)`=126.38
Now, substitute the values into the quadratic formula:

t=
`(-118+-\126.38 )/(2.16)`

`t_(1)` =
`(-118+126.38 )/(2.16)` = 8.28 seconds

`t_(2)` =
`(-118-126.38 )/(2.16)` = -7.69 seconds

The projectile reaches 112 feet at 8.28 seconds in the upward motion and in 7.69 seconds in downward motion