A ball is thrown upward with an initial velocity of 11 meters per second. Suppose that the initial height above the ground is 6 meters. At what time will the ball hit the ground? The ball is on the ground when s=0. Use the equation s=-5t^2+vt+h

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(6,1)(5,4)slope = (4 - 1) / (5 - 6) = 3/-1 = -3y = mx + bslope(m) = -3use either of ur , = 6 and y = 1now sub and find b, the y int1 = -3(6) + b1 = -18 + b1 + 18 = b19 = bso ur equation is : y = -3x + we need it in standard formy = -3x + 3x to both sides3x + y = 19 < == standard form