Question

The formula S = -16t^2 + V0t + s0 tells the height at time t of a projectile fired vertically from height s0 with initial velocity v0.

When, if ever will the projectile's height exceed 10,000 feet.
one sec, retyping equation

Answer 1

find veterx to see if it reached 10,000

Answer 2

\[S = -16t^2 + v0t + s0 \]
the 0 stands for a tiny subscript o

Answer 3

t = time, v0 = initial velocity, and s0 = height

Answer 4

let initial height be 0: \[10000 = -16t^2 +v _{0}t\]

Answer 5

ok

Answer 6

solve when discriminant >=0

Answer 7

if initial height is 0; id think about subtracting 10000 from the original height and see if anythings left above the horizontal axis

Answer 8

\[v_0^2 - 4\cdot(-16)\cdot(-10000)\ge 0\]

Answer 9

working, thank you

Answer 10

what did you get?

Answer 11

it'll never happen?

Answer 12

\[-16t^2+v_0t-10000=0\] This will have real solutions when the discriminant (b^2-4ac) is > or = to 0

Answer 13

let's think about taking the first derivative of the position function and setting it equal to zero....

Answer 14

0=-32t+Vo
32t=Vo

Answer 15

Then 10000=16t^2+32t*t

Answer 16

should be -16t^2

Answer 17

10000=16t^2

Answer 18

divide by the 16 and take the square root to find t

Answer 19

\[{v_0}^2 - 4\cdot16\cdot10000 \ge 0\]
\[{v_0}^2 \ge 4\cdot16\cdot10000 \]
\[{v_0} \ge \sqrt{4\cdot16\cdot10000} \]
\[{v_0}^2 \ge 2\cdot4\cdot100= 800{ft \over sec}\]

Answer 20

Ok, i'm slightly lost. Probably because I'm not used to dealing with subscripts and they're confusing me.

Answer 21

I'll draw it out....

Answer 22

thank you for the patience.

Answer 23

sorry, that's \[{v_0} \ge 800{ft \over sec} \]