physics
posted by courtney .
A ball is thrown vertically upwards with an initial velocity of 20.96 m/s. Neglecting air resistance, how long is the ball in the air? What is the greatest height reached by the ball?
this problem seems easy enough but I can't seem to get the answer. How can I get time using only velocity?

Two equations for constant acceleration:
v = Vo + a t
x = Xo + Vo t +(1/2) a t^2
here
a = 9.8 m/s^2
Vo = 20.96
Xo = 0
when v = 0, we are at the top
solve for t at the top (t at the ground again will be twice that by the way)
Then use that t at the top in the second equation
x = 0 + 20.96 t 4.9 t^2
to get the height at the top
if you did not belive me about doubling that t at the top to get t when it hits the ground again, there are two ways to proceed.
Way 1
let the ball drop with zero initial velocity from the top (you will see why the time falling is the same as the time rising)
Way 2
solve the second equation for t when x= 0,
of course t = 0 and t = something which I hope is t = twice t at the top should be solutions to the quadratic.
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