(t-5)*(t-1) = 0
t = 1 second (going up)
t = 5 seconts (coming back down)
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b) After how many seconds will it return to the ground?
Same equation, equals 0
solve for t
t = 0 is the time at launch
The 2nd value of t is at impact.A projectile is launched straight up from the ground level with an initial velocity of 96 ft per second. Its height in feet, "s", after "t" seconds is given by the equation: s=16t^2+96t
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It's -16t^2 + 96t
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a) At what time(s) will it reach a height of 80 feet?
-16t^2 + 96t = 80
Solve for t
-16t^2 + 96t = 80
t^2 - 6t + 5 = 0
