Answer:
The value of b is 55
The value of c is 28
The expression for the height of the ball at time t is h(t) = -16t² + 55t + 28
Explanation:
* Lets discus the situations in the problem
- The ball was lunched up from a tower of height 28 feet above
the ground
∴ The initial height = 28 feet ⇒ (initial means time is zero)
- The ball launched with an initial velocity 55 feet/second
∴ The initial velocity = 55 feet/second ⇒ (at t = 0)
- The equation of the height of the ball after t second is:
h = ut + 1/2 at² + hi, where u is the initial velocity, a is the acceleration of
gravity, t is the number of seconds since the ball was lunched
hi is the initial height which the ball was lunched from it and h is the
height above the ground
∵ The ball was lunched up
∴ The gravity acceleration is -16 feet/second²
∵ The initial velocity is upward
∴ u = 55 feet/second
∵ The tower is 28 feet above the ground
∴ hi = 28 feet
- Substitute these values in the equation
∴ h = 55t + 1/2(-32)t² + 28 ⇒ simplify
∴ h = 55t - 16t² + 28 ⇒ start with t²
∴ h = -16t² + 55t + 28
* Now compare between the two equation
∵ h = -16t² + bt + c
∵ h = -16t² + 55t + 28
∴ b = 55 ⇒ coefficient of t
∴ c = 28
∴ h(t) = -16t² + 55t + 28