A ball is projected into the air. Its height at time, t, is given by the equation h = -16t2 + 50t + 2. When will the height be 8 feet for the second time?

Question

The answers are:

A. 3.5 sec

B. 0.04 sec

C. 6.2 sec

D. 3 sec

Answer 1

h(t) = -16t^2 +50t + 2

h(3) = -16(3)^2 +50(3) +2

= -16(9) +150 +2

= -144 +150 +2

= 6+2

= 8 feet

when t= 3 seconds

h(3) = 8 feet

h(3.5) = -16(3.5)^2 +50(3.5) +2 = -16(12.25)+50(3.5)+2 = -196+175+2 =-19 feet

h(.04) = -16(.04)^2 +50(.04) +2 = -.0256 + 2+2 = 3.9744 feet

h(6.2) =-16(6.2)^2 +50(6.2) +2 = -616.04 + 310+2 =-304.04 feet

after 3.5 seconds it's hit ground and gone deeper

at .04 second, it's gone less than 2 feet up from where it started

h(.125) = h(1/8) = -16(1/8)^2 +50(1/8)+2 = -1/4 +25/4 +2 =24/4+2 =6+2 = 8 feet on the way up

at both 1/8 of a second and at 3 seconds, the ball is 8 feet high

it reaches a maximum height when h'(t)=0 = -32t +50, t=50/32=25/16 = 1 9/16 seconds

it reaches 8 feet at both 1/8 and 3 seconds. 3- 1 9/16 = 1 7/16 = 1 9/16-1/8

it reaches 8 feet at 1 7/16th of a second before and after reaching maximum height

Answer 2

8 = -16t² + 50t + 2

0 = -16t² + 50t - 6

0 = 2(8t² - 25t + 3)

0 = 2(8t + 1)(t - 3)

t = 3

Answer 3

Set 8 equal to the equation so that it can be accounted for.

8 = -16t^2 + 50t + 2

Subtract 8 from both sides so it can be set equal to 0.

0 = -16t^2 + 50t - 6

Set everything equal to 0 and use the quadratic equation

a = -16, b= 50, c= -6

X= -b +/- (√b^2- 4ac)/2a

X= -50+/-(√50^2 -4*-16*-6)/2*-16

Use order of operations PEMDAS to simplify.

You will have 2 roots.

The value that is positive will be the value that you will be going with.

The key to using the quadratic formula is evaluating it successful with no errors and watching your signs