A ball is thrown vertically upward from the ground. Its distance in feet from the ground in t seconds is s=-16t^2+128t. After how many seconds will the ball be 128 feet from the ground?

Question

campbell_st · Answer

well set s = 128 and then solve for t

so you have 

$$128 = -16t^2 + 128t$$

or re-writign the equation you have 

$$16t^2 - 128t + 128 = 0$$

divide each term by 16 and you get 

$$t^2 - 8t + 8 = 0$$

so it looks like you either need the general quadratic formula or completed the square to solve for t... 

hope it helps

you may have two values for the time (t).... one for going up... and then one for coming back down...