Question

A ball is thrown from the top row of seats in a stadium. The function h(t) = −16t2 + 64t + 80 gives the height, h, in feet, of the ball t seconds after it is thrown. How long will it be before the ball hits the ground?

Answer 1

Answer:

Step-by-step explanation:

Given the height of the ball expressed as:

h(t) = −16t^2 + 64t + 80

The ball hits the ground the when h(t) = 0

Substitute h(t) = 0 and find t as shown:

h(t) = −16t^2 + 64t + 80

0 = −16t^2 + 64t + 80

−16t^2 + 64t + 80 = 0

Divide throigh by -16

t^2-4t-5 = 0

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

t(t-5)+1(t-5) = 0

(t+1)(t-5)= 0

t+1 = 0 and t - 5 = 0

t = -1 and t = 5

time can not be negative:

t = 5seconds

Hence it will take 5 secs before the ball his the ground

Answer 2

Solving a quadratic equation, it is found that the ball hits the ground after 5 seconds.

The height of a ball after t seconds is given by:

$h(t) = -16t^2 + 64t + 80$

It hits the ground at the values of t for which:

$h(t) = 0$

Then

$-16t^2 + 64t + 80 = 0$

Simplifying by -16:

$t^2 - 4t - 5 = 0$

Which is a quadratic equation with coefficients $a = 1, b = -4, c = -5$.

Then:

$\Delta = (-4)^2 - 4(1)(-5) = 36$

$x_{1} = \frac{-(-4) + \sqrt{36}}{2} = 5$

$x_{2} = \frac{-(-4) - \sqrt{36}}{2} = -1$  

We want the positive value, thus, the ball hits the ground after 5 seconds.

A similar problem is given at https://brainly.com/question/24764843