The height of a swimmer's dive off a 10-foot platform into a diving pool is modeled by the equation y= 2x^2 - 12x + 10, where x represents the number of seconds since the swimmer left the diving board and y represents the number of feet above or below the water's surface. What is the farthest depth below the water's surface that the swimmer will reach?

a) 6 ft**
b) 8 ft
c) 10 ft
d) 12 ft

Ramon is standing on a balcony 84 feet above the ground and throws a penny straight down with an initial velocity of 10 feet per second. An equation that models the height, h(t), above the water, in feet, of the diver in time elapsed, t, in seconds, is h(t) = -16t^2 - 10t + 84. In how many seconds will the penny hit the ground?
(This is how the question was worded, I don't really understand the mentioning of a diver in this question but this is what it said)
a) 8 sec**
b) 6 sec
c) 2 sec
d) 3 sec

6 answers

y= 2x^2 - 12x + 10
This is a parabola with the vertex at the bottom
So, if you do calculus we are almost finished but with algebra complete the square or solve for zeros and look halfway between.
completing the square:
x^2 - 6 x = y/2 -5
x^2 - 6 x + 9 = y/2 +4 = (1/2) (y+8)
(x-3)^2 = (1/2)(y+8)
vertex at x = 3 and y = -8 (your bottom of the trip)
using calculus
dy/dx =0at minimum
0 = 4 x - 12
x = 3
so y = 18 - 36 + 10 = -8 again
So for the first question, it would be b) 8ft?
whoever wrote the first question wrote thee second one and the equation works for divers or pennies even if it is stuck in feet instead of meters so g = 32 ft/s^2
h(t) = initial height + initial velocity * time + (1/2)gt^2
so
h = -16t^2 - 10t + 84
at ground, h = 0
so
16 t^2 + 10 t - 84 = 0
8 t^2 + 5 t - 42 = 0
t = [ -5 +/- sqrt (25 + 1344) ] / 16
= [ -5 +/- 37 ] / 16
= 32/16 = 2 seconds
yes, 8 feet below surface of the pool
Thank you sm! This helps a lot!
You are welcome.