A ball is thrown vertically upward at an initial velocity of 32fps. The distance s (in feet) of the ball from the ground after t seconds is s=32t-16t^2. A) At what time t will the ball strike the ground? B) For what time t is the ball more than 12 feet above the ground?

Question

cwrw238 · Answer

when ball strikes the ground s = 0
0 = 32t - 16t^2

one value of t will be 0 which corresponds to the time the ball was thrown up

cwrw238 · Answer

the other value of t is what you want

anonymous · Answer

t=8?

cwrw238 · Answer

16t^2 - 32t = 0
16t(t - 2) = 0

t = 0  or t =  2

cwrw238 · Answer

2 seconds

anonymous · Answer

so if im going to figure out when its above 12 ft how would I plug it into my equation?

cwrw238 · Answer

i'm looking at this again - 2 seconds seem a very short time

cwrw238 · Answer

- no its is correct

cwrw238 · Answer

well you can first work out the time taken to get to 12 feet

from 12 = 32t - 16t^2
16t^2- 32t + 12 = 0
4t^2 - 8t + 3 = 0
t = 0.5 ,1.5

0.5 secs as it climbs  then 1.5 secs on the way down

cwrw238 · Answer

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