A baseball is thrown straight upward with an initial speed of 128 ft/sec. The number of feet s above the groun?

1)

that means the position is at s = 192

192 = -16t^2 + 128t

0 = -16t^2 + 128t - 192 ---> divide all by -16

0 = t^2 - 8t + 12

0 = (t - 6)(t - 2) <------> t = 2 & 6 seconds

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2)

to hit the ground means, that when s = 0

0 = -16t^2 + 128t

0 = 16t^2 - 128t

0 = 16t * (t - 8)

16t = 0 -----> t = 0 seconds

t - 8 = 0 ----> t = 8 seconds

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you have your preliminary velocity 20m/s. on the suitable of its direction, the stone would have a velocity of 0 m/s. Acceleration by gravity would desire to be -9.8 m/s^2 (destructive thinking the shown fact that acceleration precise it particularly is destructive to the path of action). utilising the third equation of action v^2=u^2+2aS, with v as very final velocity, u as preliminary velocity, a as acceleration by gravity for that reason and S as a results of fact the suitable attained by ability of utilising the stone. So, after changing, the equation would desire to look as though this: 0^2 = 20^2 - 2*9.8*S fixing for S, you will get 20.4m, which the optimal suitable the stone would desire to be triumphant in.