Derivative Application

Question

An object is launched straight upward from a platform. Its height above ground is

h(t) = h₀+ v₀t − 16.1t²

where h is in feet and t is in seconds. The graph of h is shown below.

Find both unknown constants, accurate to 1 decimal place.

Answer 1

h(t) = ho +vot -16.1t^2, ho= initial height at time t=0

velocity = v(t)=h'(t) = vo -32.2t = 0 at maximum height when h= 126 feet

t=vo/32.2, vo= initial velocity at time t=0

2.5 = vo/32.2

vo = 2.5(32.2) = 64.4+16.1 = 80.5

h(t) = ho + 80.5t - 16.1t^2

126 = h(2.5)= ho +80.5(2.5) - 16.1(2.5)^2

126=ho +201.25 - 16.1(6.25)

ho = 126 - 201.25+ 100.625

ho =226.625 -201.25 = 25.375

ho = 25.375 = 25.4 feet

h(t) = 25.4 + 80.5t -16.1t^2

normally this type of problem uses -16t^2 not -16.1t^2. It might be a misprint or typo

it reflects gravity which is -32 feet per second per second at sea level, without air resistance,

as if in a vacuum. a(t)=v'(t)=h"(t)=-32 ft/sec^2

using the conventional -16t^2 gives a slightly different answer, which seems more likely

h'(t)=v(t)=vo-32t = 0

t = 2.5=vo/32

vo=80 ft/sec

126=ho+80(2.5)-16(2.5)^2

126=ho+200-16(6.25)

ho=26

the numbers come out evenly to integers suggesting -16.1t^2 was a misprint or miscopied

h(t) =26+80t-16t^2

also the position of the y intercept on the graph visually looks closer to 27 or 26, than 25.4.

each of the 7 increments on the vertical axis = 126/7 = 18

the y intercept is about one and one-half increments = 1.5(18) = about 27.

the graph is a downward opening parabola

with axis of symmetry x=2.5

another point on the parabola is (1,90) after 1 second it is 90 feet high

(1,90) is the clearest point, go to 1 second on the x axis, then up 5 squares, it intersects the graph at

(1,5x18) = (1,90)

h(1) = 90 = ho + 80(1)- 16(1^2)

ho =90-80+16 = 26 feet initially

another method is take the vertex (maximum) off the graph (2.5, 126)

and plug it into the quadratic in vertex form

a(t-h)^2+ k where (h,k)= vertex, a=-16

=-16(t-2.5)^2 +126

=-16(t^2- 5t + 6.25)+126

=-16t^2+80t -100+126

= -16t^2+80t+26 with 80=vo, 26=ho

but if you really want to stick with -16.1t^2 then with the same vertex form:

-16.1(t-2.5)^2 +126

-16.1(t^2-5t +6.25)+126

-16.1t^2 + 80.5t-100,625 +126

-16.1t^2+80.5t + 25.375

then vo = 80.5 ft/sec, ho =about 25.4 feet high = the platform's height

but then

f(1) =-16.1+80.5+25.375 =105.375-16.1 = 88.625 but the graph shows that point as (1,90) not (1,88.625)

suggesting there was a typo, misprint or miscopy and -16t^2 was correct

then vo=80 ft/sec and ho= 26 ft

this problem might have been listed under physics as well as calculus

the more accurate gravity acceleration constant is 32.1740 at sea level with no air resistance

the problem uses rounding to 1 decimal place. 32.1740/2 = 16.0870 rounded to one decimal= 16.1

so there's the counter argument that 16.1 is correct and not 16.

But the platform is above sea level, and with any air resistance, the gravity constant looks closer to 16.0 than to 16.1

Answer 2

We know that at t = 2.5 d/dt = 0 because the tangent line is horizontal. Taking the derivative we get v0 = 32.2t. t = 2.5 so v0 = 80.5 ft/s. Now we can plug this in to find h(0). 126 = h(0) + 80.5(2.5) - 16.1(2.5^2). Solving this equation for h(0) we get h(0) = 25.375 ft. So in short v(0) = 80.5 and h(0) = 25.4.