Question

Arthur drops a ball from a height of 81 feet above the ground. Its height, h, is given by the equation h = –16t^2 + 81, where t is the time in seconds. For which interval of time is the height of the ball less than 17 feet?

Answer 1

Step-by-step explanation:

Arthur drops a ball from a height of 81 feet above the ground. Its height as a function of time is given by :

$h(t)=-16t^2+81$...........(1)

Where

t is in seconds

We need to find the time interval when the height of the ball is less than 17 feets. Equation (1) becomes :

$-16t^2+81<17$

$-16t^2<17-81$

$-16t^2<-64$

$t^2>4$

$t>2$

When the time is more than 2 seconds, the height of the ball is less than 17 feet. Hence, this is the required solution.

Answer 2

Answer:A

Step-by-step explanation:

On edge