h(t) = -16t^2 + 40t + 5 represents the height, in feet, of an object above level ground at time t, in seconds

Question

Given this information, complete the following sentence with one of the choices listed below.

The object will hit the ground when t = ______ seconds

A. 0.9
B. 1.8
C. 2.6
D. 3.2
E. none of these

Answer 1

h(t) = -16t^2 + 40t + 5

x = (-b ± sqrt(b^2 - 4ac))/2a

x = (-40 ± sqrt((40)^2 - 4(-16)(5)))/(2(-16))
x = (-40 ± sqrt(1600 + 320))/-32
x = (-40 ± sqrt(1920))/(-32)
x = (-40 ± sqrt(64 * 30))/(-32)
x = (-40 ± 8sqrt(30))/(-32)
x = (5/4) ± (1/4)sqrt(30)
x = about 2.619

ANS : C.) 2.6 seconds

Answer 2

So the question is, when does this equation equal zero?

Test the cases:

if t = 0.9, h(t) = (-16)(0.9)^2 + 40(0.9) + 5 = -16(0.81) + 36 + 5 = -12.96 + 41 = 28.04
if t = 1.8, h(t) = (-16)(1.8)^2 + 40(1.8) + 5 = -16(3.24) + 72 + 5 = -51.84 + 77 = 25.16
if t = 2.6, h(t) = (-16)(2.6)^2 + 40(2.6) + 5 = -16(6.76) + 104 + 5 = -108.16 + 109 = 0.84

While it's not EXACTLY 2.6 seconds, it's within a couple of hundredths of a second of that mark. That's your best answer.

Answer 3

nicely, at t = 0 it changed into at 5 ft. At t = 2 that is at h = -sixty 4 + eighty + 5 = 21 ft (only checking to verify it hasn't hit the floor already...if it had, the answer may be E regardless of the favor to tell apart the equation some situations!!) in case you differentiate the equation: dh/dt = -32t + 40 = v evaluate at t = 2 seconds supplies v = dh/dt = -24 ft/sec (ie - falling as a results of detrimental signal) Differentiate again: dv/dt = a = -32 ft/sec/sec (Earth gravitational acceleration) that is accelerating interior a similar direction (signal) as its speed at t = 2 seconds, therefore that is C (falling at an increasing cost).

Answer 4

just try them out when the thing hits the ground h(t) =0

Answer 5

you think too much. i'd like to see how many people know the correct answer

Answer 6

C. 2.6

with some rounding...

Answer 7

c

Answer 8

2.6