Question

A ball is thrown from a height of 38 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= 38-3t-5t^2

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

2.47 seconds

Explanation:

The height is zero when the ball hits the ground, so time can be found by solving the quadratic ...

-5t^2 -3t +38 = 0

for t in any of the usual ways. A graphing calculator gives a solution quickly. The quadratic formula can give you the "exact" solution.

For that formula, we have a=-5, b-3, c=38, so the solution looks like ...

t = (-b ±√(b^2-4ac))/(2a)

t = (-(-3) ±√((-3)^2 -4(-5)(38)))/(2(-5))

= (3 ±√769)/(-10)

We're only interested in the positive solution, so ...

t = -0.3 +√7.69 ≈ 2.473085

The ball hits the ground about 2.47 seconds after it is thrown.

Answer 2

Hello!

The answer is:

The ball will hit the ground after
`t=2.47s`

Why?

Since we are given a quadratic function, we can calculate the roots (zeroes) using the quadratic formula. We must take into consideration that we are talking about time, it means that we should only consider positive values.

So,

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

We are given the function:

`h=-5t^(2)-3t+38`

Where,

`a=-5\\b=-3\\c=38`

Then, substituting it into the quadratic equation, we have:

`\frac{-b+-\sqrt{b^(2) -4ac} }{2a}=\frac{-(-3)+-\sqrt{(-3)^(2) -4*-5*38} }{2*-5}\\\\\frac{-(-3)+-\sqrt{(-3)^(2) -4*-5*38} }{2*-5}=(3+-√(9+760) )/(-10)\\\\(3+-√(9+760) )/(-10)=(3+-√(769) )/(-10)=(3+-27.731)/(-10)\\\\t1=(3+27.731)/(-10)=-3.073s\\\\t2=(3-27.731)/(-10)=2.473s`

Since negative time does not exists, the ball will hit the ground after:

`t=2.473s=2.47s`

Have a nice day!