SOLUTION: An object is shot straight up into the air from ground level with initial speed of 24 ft/sec. The height of the object (in feet) is given by the equation h=-16tsquared+24t, where t

Algebra ->  Algebra  -> Equations -> SOLUTION: An object is shot straight up into the air from ground level with initial speed of 24 ft/sec. The height of the object (in feet) is given by the equation h=-16tsquared+24t, where t      Log On

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 Question 79507: An object is shot straight up into the air from ground level with initial speed of 24 ft/sec. The height of the object (in feet) is given by the equation h=-16tsquared+24t, where t is the time in seconds after launch. Find the time(s) when the object is at ground level.Answer by stanbon(57395)   (Show Source): You can put this solution on YOUR website!An object is shot straight up into the air from ground level with initial speed of 24 ft/sec. The height of the object (in feet) is given by the equation h=-16tsquared+24t, where t is the time in seconds after launch. Find the time(s) when the object is at ground level. --------- At ground level the height is zero. To get the time let h=0 and solve for t: -16t^2+24t=0 -8(2t^2-3t)=0 2t^2-3t=0 t(2t-3)=0 t=0 or t=3/2 This means the object was on the ground to start with (t=0) and will be on the ground again in 1.5 seconds (t=3/2) ========== Cheers, Stan H.