Posted by **courtney** on Wednesday, January 16, 2008 at 6:17pm.

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?

- physics -
**Damon**, Wednesday, January 16, 2008 at 7:29pm
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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