one more really fast.?

Question

A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s. When will the ball reach a height of 80 m?

Answer 1

Not accounting for deccelration due to gravity or wind resistance? 4 seconds. Where X=Time:

20X=80
X=80/20
X=4

Answer 2

There is no way to avoid gravity!
Anyway, make an equation:
x = x0 + v0*t + 1/2*a*t^2
That means, in words:
position = initial position + initial velocity times time + one-half acceleration times time-squared
You'll notice that if you're solving for t, it will be a quadratic. Look at the values you already have:
x = 80 (height in question)
x0 = 100 (launching height)
v0 = 20 m/s (launching velocity)
t = ?
a = -9.8 m/s^2 (gravitational acceleration constant)
t, again = ?

Plug 'em in.
80 = 100 + 20t - 4.9t^2

Solve for t.
t = -.831 or t= 4.9124
It obviously isn't negative.
So

t = 4.912 seconds

Answer 3

4 seconds
if it is going 20 m/s and it wants to go 80 m you divide 80 by 20 and get 4

Answer 4

use the equation y=yo+vot+1/2at^2
where yo=initial height, vo=initial v, a=g=-9.8m/s^2

so 80=100+20t-4.9t^2
-4.9t^2+20t+20=0
graph this use the quadratic formula to solve for t, one of the values will be obviously wrong so use the positive value of t

Answer 5

a = dv/dt = -g
v = -gt +C
When t= 0 v = 20, so
v = -gt +20
s = -gt^2/2 +20t +C
When t = 0 s = 100 ,so
s = -gt^2/2 +20t +100
s= -4.9 t^2 +20 t +100 , since g = -9.8 m/sec^2

80 = -4.9t^2 +20t +100
0 = -4.9t^2 +20t+20
t = [ -20 +/- sqrt(20^2 -4(-4.9)(20)]/(2*(-4.9))
t = [-20 +/- sqrt(792)]/(-9.8)]
t= [-20 +/- 28.14]/-9.8
t = -48.14/-9.8 = 4.91 seconds