Q:

Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 32feet above the ground, the function h(t)=−16t2+48t+32 models the height, h, of the ball above the ground as a function of time, t. At what time will the ball reach a height of 64feet?

Accepted Solution

A:
Answer:There are two times for the ball to reach a height of 64 feet:1 second after thrown ⇒ the ball moves upward2 seconds after thrown ⇒ the ball moves downwardStep-by-step explanation:* Lets explain the function to solve the problem- h(t) models the height of the ball above the ground as a function  of the time t- h(t) = -16t² + 48t + 32- Where h(t) is the height of the ball from the ground after t seconds- The ball is thrown upward with initial velocity 48 feet/second- The ball is thrown from height 32 feet above the ground- The acceleration of the gravity is -32 feet/sec²- To find the time when the height of the ball is above the ground   by 64 feet substitute h by 64∵ h(t) = -16t² + 48t + 32∵ h = 64∴ 64 = -16t² + 48t + 32 ⇒ subtract 64 from both sides∴ 0 = -16t² + 48t - 32 ⇒ multiply the both sides by -1∴ 16t² - 48t + 32 = 0 ⇒ divide both sides by 16 because all terms have   16 as a common factor∴ t² - 3t + 2 = 0 ⇒ factorize it∴ (t - 2)(t - 1) = 0- Equate each bracket by zero to find t∴ t - 2 = 0 ⇒ add 2 to both sides∴ t = 2- OR∴ t - 1 = 0 ⇒ add 1 to both sides∴ t = 1- That means the ball will be at height 64 feet after 1 second when it  moves up and again at height 64 feet after 2 seconds when it  moves down* There are two times for the ball to reach a height of 64 feet   1 second after thrown ⇒ the ball moves upward   2 seconds after thrown ⇒ the ball moves downward