Question

A ball is thrown from a height of 64 meters with an initial downward velocity of 3 m/s. The ball's height h (in meters) after t seconds is given by the

following
h=64-31-51?
How long after the ball is thrown does it hit the ground?
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)

Answer 1

The ball lands in about 3.29 seconds.

The given equation is:

`h=-5t^2-3t+64`.

Explanation:

I do believe the equation is meant to be:

`h=-5t^2-3t+64`.

We want to find
`t` such that
`h=0` since we want to find the time when the ball has hit the ground.

`0=-5t^2-3t+64`

Our objective here is to solve this equation using the quadratic formula,
`t=(-b\pm √(b^2-4ac))/(2a)`.

First we need to compare the equation we have to
`0=at^2+bt+c`.

`a=-5`

`b=-3`

`c=64`

`t=(-(-3)\pm √((-3)^2-4(-5)(64)))/(2(-5))`

Simplify the denominator first:

`t=(-(-3)\pm √((-3)^2-4(-5)(64)))/(-10)`

Simplify the inside of the square root:

`t=(-(-3)\pm √(1289))/(-10)`

Simplify the -(-3):

`t=(3\pm √(1289))/(-10)`

So we have either
`t=(3+√(1289))/(-10)` or
`t=(3-√(1289))/(-10)`.

Let's both of those into our calculator.

`t=-3.89` or
`t=3.29`

So the ball lands in about 3.29 seconds.