A ball be thrown vertically from ground level and rise to a maximum height of 80 m?

Question

How long will it be in the air?

Answer 1

Well to reach a max height of 80m you will need to know the initial velocity which can be found with

V^2 = Vo^2 + 2a*delta X

Solving for Vo you get 39.6 m/s

Then you would plug that into

X = Xo + Vo*t + 1/2*a*t^2

Simplifying gives that t = 2*Vo/ a

Plugging all the values in you get a t of 8.08 seconds

Answer 2

s = distance traveled
v0 = initial velocity
g = acceleration of gravity (acting downwards) = 9.8 m/s^2
t = the time to reach maximum altitude

s = v0t - (1/2)gt^2 = 80

Magnitude of the initial velocity is the same as that of the velocity at impact.
v0 = gt

Put these together and get:
(1/2)gt^2 = 80
gt^2 = 160

t =4.04 seconds

Total time in the air will be twice this (trajectory is symmetrical, same time coming down as going up)

Total time = 2(4.04) = 8.08 seconds

Answer 3

We will ignore air resistance.
The equation of motion is
y(t)=y0+v0*t-.5*g*t^2
ymax=80
y0=0
v0=unknown
use g=10
t= unknown
The velocity equation is
v(t)=v0-g*t
the v at ymax = 0
so
80=v0*t-5*t^2
and
0=v0-10*t
where t is the time to apogee, or max height
combining the equations
v0=10*t
and
80=10*t^2-5*t^2
80/5=t^2
t=sqrt(16)
t=4 seconds

I already know that the total flight time is 8 seconds, but I will derive it.
Knowing the time to ymax is 4, then
v0=40 m/s
also,
y(t)=0 at takeoff and landing so
0=40*t-5*t^2
note that t=0 is one correct solution
the other root of this quadratic is
t=8

j

Answer 4

8.08 seconds.