Question

If a baseball player hits a baseball from 4 feet off the ground with an initial velocity of 64 feet per second, how long will it take the baseball to hit the ground? Use the equation `h = –16t^2 + 64t + 4`?

Answer 1

`t=+4.06` to 2 decimal places

Explanation:

This is a quadratic in `t` instead of `x`. It behaves in exactly the same way as it would if we had an `x` instead of a `t`

Using the formula

`color(red)("Set "h=0)=y=ax^2+bx+c => x=(-b+-sqrt(b^2-4ac))/(2a)`

Where `a=-16; b=+64 and c=+4`

Set `x=t` giving:

`=>t=(-64+-sqrt((-64)^2-4(-16)(+4)))/(2(-16))`

`t=+2+-sqrt(+4352)/(-32)`

`t=-0.6155... and t=+4.06155...`

The negative solution is not applicable so we have:

`t=+4.06` to 2 decimal places

`color(red)("Always state the number of places rounded to")`

Answer 2

`4.06s` (3s.f.)

Explanation:

when the baseball hits the ground, the height will be `0`.

`h = 0`

we are given that `h = -16t^2 + 64t + 4`.

this means that the height is `0` when `-16t^2 + 64t + 4 = 0`.

`-16t^2+64t + 4 = 0`

divide both sides by `4:`

`-4t^2 + 16t + 1 = 0/4 = 0`

`-4t^2 + 16t + 1 = 0` can be solved using the quadratic formula:

`t = (-b +- sqrt(b^2-4ac))/(2a)`, where `at^2 + bt + c = 0`.

`at^2+bt+c = -4t^2+16t+1`

`a = -4, b = 16, c = 1`

`t = (-b + sqrt(b^2-4ac))/(2a) or (-b - sqrt(b^2-4ac))/(2a)`

`(-b + sqrt(b^2-4ac))/(2a) = (-16+sqrt272)/(-8)`

`=-0.06..`

`(-b - sqrt(b^2-4ac))/(2a) = (-16-sqrt272)/(-8)`

`=4.06` (3s.f.)

we have `2` values for `t: -0.06` and `4.06`.

since time cannot be negative, only the positive value can be taken.

this is `4.06`, to `3` significant figures.

the time taken is `4.06` seconds.