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
