Polynomial functions graphs applications and model question

Question

A toy rocket is launched from the top of a building 80 feet tall and initial velocity of 219 ft./s.

A. Give the function that describes the height of the rocket in terms of time t.

B. Determine the time at which the rocket reaches its maximum height, and the maximum height in feet.

C. For what time interval will the rocket be more than 546 feet above ground level?

D. After how many seconds will it hit the ground?

Answer 1

Part A:

h(t) = -16t^2 + Vo t + Ho

where -16 is the gravity constant, Vo is the

intital velocity, and Ho is the initial height

at time t=0

Vo=219 and Ho=80

h(t) = -16t^2 + 219t + 80

Part B:

h(t) = -16t^2 + 219t + 80

A=-16 B= 219 C = 80

tMax= -B/(2A) = -219/(2*-16) = -219/-32 = 6.84375

h(6.84375) = 829.390625 is the max height

Part C:

-16t^2 + 219t + 80 > 546

-16t^2 + 219t - 466 > 0 <--- subtracts 546

16t^2 - 219t + 455 < 0 <--- multiplies or divides by -1, changes the sign

If equal to zero, the quadratic formula says:

t = [219 +or- sqrt ( (-219)^2 - 4(16)(455))] / 32

= [219 +or- sqrt(18841)]/32

= [219 +or- sqrt(18841)]/32

So the rocket is beyond 546 feet when [219 - sqrt(18841)]/32 < t < [219 + sqrt(18841)]/32

sqrt(18841) is APPROXIMATELY 137.262522197430133044697162733

so the time frame is APPROXIMATELY 2.55+ seconds to 11.13+ seconds

Part D:

-16 t^2 + 219t + 80 = 0

16t^2 -219t - 80 = 0

t = [219 +or- sqrt( 219^2 - 4(16)(-80)] / 32

= [219 +or- sqrt( 219^2 - 4(16)(-80)] / 32

The negative branch results in negative measures

t = [219 + sqrt(53081)]/32

which is APPROXIMATELY 14.04353+ seconds