A circus acrobat is shot out of a cannon from a platform 25 feet high. The equation for the acrobat’s pathway can be modeled by: h(t) = -16t² + 80t + 25.

Question

A circus acrobat is shot out of a cannon from a platform 25 feet high. The equation for the acrobat’s pathway can be modeled by:  

h(t) =  -16t² + 80t + 25. 

a. Find the maximum height of the acrobat. 

b. Write the equation that model’s the acrobat’s pathway in Vertex Form.   

c. What is the rate of change on the interval 3 ≤ x ≤ 5.

Answer 1

max height = 125 feet;

vertex form h(t) = -16(t-2.5)^2 + 125

average rate of change from 3 to 5 is -48 feet per second

take the derivative and set = 0, then solve for t

h'(t) = -32t +80 = 0

32t=80,

t=80/32 = 2.5 seconds to reach maximum height

plug that value of t into the original equation

-16(2.5)^2 +80(2.5) + 25 = -100 + 200 +25= 125 feet = max height

-16t^2 +80t +25 factor out -16 from x and x^2 terms

(-16)(t^2 - 5t ) +25

complete the square, take half the coefficient of the t term and square it,

add it within the parentheses, which is the same as subtracting 16(25/4)=100

so add also add 100

(-16)(t-5t + 25/4) + 25 +100

-16(t-2.5)^2 + 125

f(t)= (-16)(t-2.5)^2 +125 in vertex form

f(t) =a(t-h)^2 + k where (h,k) = vertex

vertex is (2.5, 125)

rate of change = velocity = v = derivative of h(t)

h'(t) = v(t) = -32t + 80

v(3) = -32(3) + 80 = 80-96 = -16 ft/sec

v(5) =-32(5) + 80 = 80- 160 =-80 ft/sec

average rate of change =[v(3)+v(5)]/2 = (-16-80)/2 = -96/2 =- 48 ft/sec

It's negative since it's falling, having passed the max height

average rate of change over an interval is the change in h over the change in t

change in h = h(5)-h(3) = -16(5)^2+80(5)+25 = -400+400+25 - (-16(3)^2+80(3)+25) = 25 -25 +32(9)-240 = 144-240 = -96 at time t=3 its reached back to the height it started. at time t=5, it's fallen further

change in t = 5-3=2

change in h over change in t = -96/2 = -48 feet per second

Graphically, it's a downward opening parabola, with maximum point or vertex at (2.5, 125), y intercept at (0,25), x intercepts about (-1/4, 0) and (5 1/4, 0),

axis of symmetry x= 2.5