Question

A ball is dropped from a height of 784 ft. It's height h, in feet, after t seconds is given by h(t)=784-16t^2. After how long will the ball reach the ground?

Answer 1

`t= 7 sec`

Explanation:

So here height is given as function of time, and I want to know when the height is equal to 0, so set it equal to 0:

`0= -16t^2+784`

The goal here is to solve for `t`:

Subtract `784` from both sides
`-16t^2=-784`

Divide by `-16`:
`t^2= 49`

Square root both sides:
`t =+-7`

Since time can't be negative:
`t= 7 sec`

graph{-16x^2+784 [-10, 10, -5, 5]}

Answer 2

`t=7`

Explanation:

Since you are trying to find when the ball hits the ground, `h(t)=0` because the ground is at `0` feet.
`h(t)=0=784-16t^2` Isolate `t` and its accompanying exponents/coefficients.
`16t^2=784` Divide by 16 to isolate `t` and its exponent.
`t^2=49` Find the square root of each side.
`t=sqrt(49)` `(7^2=49)`
`t=+ or - 7` Since `t` represents time, it must be positive, so
`t=7`

Answer 3

`7` seconds

Explanation:

The height of the ball is given by the function `h(t)=784-16t^2`.

When the ball is on the ground, we can say that its height is `0`.

And so,

`784-16t^2=0`

`16t^2=784`

`t^2=784/16=49`

`t=sqrt(49)`

`t=+-7`

But since `t` is the time, and time cannot be negative, we only take the positive root, that is:

`t=7`

So, it'll take seven seconds for the ball to hit the ground.