Question

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?

Answer 1

Answer:

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

Step-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