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