Question

satellite please explain how you got your answer from last post 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.
Suppose a projectile is fired vertically from ground level at an initial speed of 448 ft per second.
When will the projectile be higher than 3072 ft?

Answer 1

-16t^2 ....

Answer 2

mathematics doesnt tell you how to find out physics; it just helps you to express what you find

Answer 3

the acceleration due to gravity is a constant; it stays the same.

on earth that constant is -32 ft/sec/sec

Answer 4

solve for the equation\[3072\leq-16t^2+v_0t +s_0\ |_{v_0 = 448,\ \ \ s_0=0}\]

Answer 5

a(t) = -32

it just so happens that \(\int a(t)\) = velocity

v(t) = -32t + C

Answer 6

oops I mean 3072 < expression

Answer 7

ok lets start at the beginning without integration etc

Answer 8

lets call it better as:

v(t) = -32t + Vi ; for initial velocity at time = 0

another neat trick is that velocity is the derivative of height/position suc that: \(\int v(t)\) = s(t)+C

Answer 9

ok lets start at the beginning without integration etc

Answer 10

or rather:

s(t) = -16 t^2 +Vi t + C; where the initial position at time = 0 is given as C such that:

s(t) = -16 t^2 +Vi t + Hi

Answer 11

good god is this slow

Answer 12

its been like that all day, i emailed the shadow and he might be on in a few hours to see what he can do about it

Answer 13

print filter(None,map(lambda y:y*reduce(lambda x,y:x*y!=0, map(lambda x,y=y:y%x,range(2,int(pow(y,0.5)+1))),1),range(2,1000)))

Answer 14

ok you have the formula given to you above, and we see that the initial height was 0 because it was launched from the ground and the initial velocity was 448 because you are told

Answer 15

oops I mean I got 12s < t < 16s

Answer 16

so you get the formula
\[S(t)=-16t^2+448t\] and you are asked when this is greater than 3072, so put
\[-16t^2+448t>3072\] and solve

Answer 17

ok the problem is i dont know how to solve it :/

Answer 18

To solve it, use the quadratic formula, or complete the square

Answer 19

first step is
\[-16t^2+448t-3702>0\] second step is
\[-16(t-12)(t-16)>0\]

Answer 20

damn is this slow. ok almost done. the zeros are at t=12 and t=16, and it is positive between them, so that is your answer

Answer 21

positive between them because this is a parabola that faces down, so it is positive between the zeros and negative outside. there finished!

Answer 22

sorry it took so damned long but for some reason i cannot load the pages

Answer 23

thank you all so much i guess i just dont unerstand that whole parabola thing and the whole 12<x<16 or whatever it was