Acannonball was launched with an initial upward velocity of 55 feet per second from a tower 28 feet off the ground. the height of the ball above the ground, in feet, can be modeled by an expression having the form -16t^2 + bt + c, where t is the number of seconds since the cannonball was launched.

what is the value of b, the coefficient of t?

what is the value of c, the constant term?

therefore, the expression for the height of the ball at time t is

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

Step-by-step 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

b. 153 students do not participate in sports and seventy-five 12th graders were surveyed

the answer is b.

64

step-by-step explanation:

c=2πr=2·π·8≈64