math homework help please!?

Question

If a ball is thrown vertically upward from a height of 5 ft. with an initial velocity of 80 ft/sec, its height h after t seconds is given below. How long does it take the ball to return to the ground? Round to two decimal places.
h=-16t^2+80t+5

Answer 1

The function h(t) = -16 t^2 + 80 t + 5 gives
you the height of the ball at time t.

At t=0 (when you thrown upwards the ball) you have
h(0) =5.

The time T it takes to return to the ground is a solution
of the following equation

- 16 T^2 + 80 T +5 = 0

Solving this quadratic equation you get

T = 1/4(10 + SquareRoot(105))

so it's about 5.06174 seconds.

Note: There is also a negative solution

T = 1/4(10 - SquareRoot(105)) = - 0.06173 seconds

It's not unphysical. Can you guess what it means?

For Grampedo: there's a minus sign before 0,06.. ;)

Answer 2

The ball is on the ground when h = 0.
So
-16t² + 80t + 5 = 0.
16t² -80t -5 = 0.
t = 1/32*[ 80 + √(6400+320) ] = 5.06 sec(approx).
Note that we rejected the other root, because it
is negative and t represents positive time here.

Answer 3

I'm going to do this problem in 2 parts, because I am
unsure of what the question is.
Do they want the TOTAL time involved from the time
the ball is thrown upwards? I think yes.
Or do they want the time it takes to fall to the ground
once it has reached maximum height?

Part 1.
h= -16t^2+80t+5
Maximum height is reached when speed is 0.
Speed, dh/dt, = derivative of -16h^2+80t+5
dh/dt=-32t+80
At dh/dt=0, -32t+80=0
32t=80
t=2.5

Part 2.
Ball hits the ground when h =0.
-16t^2+80t+5=0
16t^2-80t-5=0
Use quadratic formula to solve.
t= {-b+/-rt(b^2-4ac)} / 2a, where
a=16, b=-80,c=-5
t={80+/-rt[(80^2-4(16)(-5)]} / 32
t={80 +/-rt(6400-320)}/ 32
t={80 +/- rt(6080)} /32
t={80+/- 77.97} /32
t=(80+77.97)/32 or (80-77.97)/32
t=157.97/32 or 2.03/32
t=4.94 or 0.06

Total time up and down=2.5 + 4.94=7.44 seconds.
Time for just the down part=4.94 seconds.

I've discounted the answer t=0.06 seconds. It may
be a solution to the math problem, but it isn't a
practical answer for the physics problem.
(Actually, I think the t=0.06 seconds is the time
needed to cover the last 5 ft.)

I hope this serves your purposes. Season's Greetings to you!

Answer 4

holy crap
well ok look up math help on the internet and you should be able to find it there