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?

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:

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

Then

Simplifying by -16:

Which is a quadratic equation with coefficients .

Then:

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

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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 are

step-by-step explanation: