Question

A ball is thrown downward from a window in a tall building. It's at time t in seconds is given by
s(t)=16t^2+32t, where s is in feet. How long will it take the ball to fall 168 ft?
Help me out please.

Answer 1

\[16t^2+32=168 \implies 2t^2+4t-21=0\]

Answer 2

Now you can use the quadratic formula to find t. Take only the positive value of t.

Answer 3

something is seriously wrong here. is this supposed to be the equation for the height of the ball? if so, it is going to go up to the heavens and beyond

Answer 4

"It's ______ at time t in seconds is given by"

Answer 5

position (srry)

Answer 6

well that is impossible.

Answer 7

first of all if it is at the top of a tall building, then its height at time t = 0 should be the height of the building, but you have
\[S(0)=0\]

Answer 8

secondly,
\[S(t)=16t^2+32t\] is a parabola that faces up, so the larger t is the greater S is, in other words as time goes on the ball rises without bound. so there is definitely something not right about this equation

Answer 9

??? what? it's not at the top of a building, it's thrown from a window in the building. i don't get what you're saying.

Answer 10

the ball isn't rising, it's going down

Answer 11

at t = 0 you have
\[S(0)=16 \times 0^2+32\times 0=0\] so its initial position is at 0

Answer 12

Good point @satellite. It probably just represents the distance it traveled t, and we restrict t between 0 and c, where c is the time at which the ball hits the ground.

Answer 13

the whole question doesn't make any sense. the parabola
\[y=16t^2+32t\] opens up and is zero at
\[t=-2,t=0\]

Answer 14

the equation should be something like
\[S(t)=-16t^2-32t+168\] if the ball starts 168 feet up and is thrown downward with initial velocity of -32 ft/sec

Answer 15

Yeah 0 at 0 works, if we consider s to be the distance traveled. Right?

Answer 16

then we could set this equation = 0 and solve for t, and see when the ball hits the ground. that would make some sense, but this question as stated doesn't

Answer 17

but i don't think it starts 168 feet up. it starts before that and we're supposed to calculate how long it would take before it falls 168 ft.

Answer 18

are you sure it is
\[16t^2\] and not
\[-16t^2\]?

Answer 19

yeah

Answer 20

If we assumed s to be the displacement, then the ball will hit ground when s equal the height of which the ball was thrown. I agree that there's something wrong with the problem as stated though!

Answer 21

ok i will let you work it out and not be cranky, but i assure you that for
t > 0 the ball hall whose position is given by
\[s(t)=16t^2+32t\] is increasing, going up, headed skyward. this function is increasing for
t > 0