SOLUTION: Jason jumped off a cliff into the ocean while vacationing with some friends. His height as a function of time could be modeled by the function h(t)= ­16tē + 16t + 480, where t is t

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Jason jumped off a cliff into the ocean while vacationing with some friends. His height as a function of time could be modeled by the function h(t)= ­16tē + 16t + 480, where t is t      Log On


   

Question 389256: Jason jumped off a cliff into the ocean while vacationing with some friends. His height as a function of time could be modeled by the function h(t)= ­16tē + 16t + 480, where t is the time in seconds and h is the height is feet.
a. How long did it take for Jason to reach his maximum height?
b. What was the highest point that Jason reached?
c. Jason hit the water after how many seconds?
d. Sketch the Problem labeling the variables involved and relevant information (y-intercept, x-intercept, vertex, axis symmetry).
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi,  

Note: the vertex form of a parabola, y=a%28x-h%29%5E2+%2Bk 

where(h,k) is the vertex 

  h(t)= ­-16tē + 16t + 480  Believe there should be a minus squared term   

h(t)= ­-16t^2 + 16t + 480  completeing the square to find the vertex

h(t)= ­-16(t^2 - t) + 480

h(t)= -­16[(t - 1/2)^2 - 1/4] + 480

h(t)= ­-16(t - 1/2)^2 + 4 + 480

h(t)= ­-16(t + 1/2)^2 + 484 vertex is Pt(1/2,484) or ordered pair (t,h(t))

a. How long did it take for Jason to reach his maximum height? 1/2  sec

b. What was the highest point that Jason reached? 484ft

c. Jason hit the water after how many seconds?

h(t)= 0 = ­­-16tē + 16t + 480

t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

t+=+%28-16+%2B-+sqrt%2830976%29%29%2F%28-32%29+

t+=+%28-16+%2B-+176%29%2F%28-32%29+

  t = 160/32 = -5 extraneous solution

  t = 192/32 = 6 sec