arthur drops a ball from a height of 81 feet above the ground. its height, h, is given by the equation h =…

arthur drops a ball from a height of 81 feet above the ground. its height, h, is given by the equation h = -16t² + 81, where t is the time in seconds. for which interval of time is the height of the ball less than 17 feet?\no t>2\no 0<t<2\no t = 2\no -2<t<2

arthur drops a ball from a height of 81 feet above the ground. its height, h, is given by the equation h = -16t² + 81, where t is the time in seconds. for which interval of time is the height of the ball less than 17 feet?\no t>2\no 0<t<2\no t = 2\no -2<t<2

Answer

Answer:

A. $t > 2$

Explanation:

Step1: Set up the inequality

We want to find when $h<17$. Given $h = - 16t^{2}+81$, so we set up the inequality $-16t^{2}+81<17$.

Step2: Rearrange the inequality

Subtract 81 from both sides: $-16t^{2}<17 - 81$, which simplifies to $-16t^{2}<-64$.

Step3: Solve for $t^{2}$

Divide both sides by - 16. Remember to reverse the inequality sign when dividing by a negative number. We get $t^{2}>4$.

Step4: Solve for $t$

Taking the square - root of both sides, we have $t > 2$ or $t<-2$. But since $t$ represents time (and time cannot be negative in this context), we discard $t < - 2$. So the solution is $t>2$.