The Chs baseball team was on the field and the batter popped the ball up. The equation b(t)=80t-16t^2+3.5. Represents the height of the ball above the ground in feet as a function of time in seconds. How long will the catcher have to get in the position to get the ball before it hits the ground?
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What we are looking for is the time when the ball hits the ground. If the ball is on the ground, then b(x) = 0. The solution for b(x) = 0:b(x) = 80t - 16t^2 + 3.50 = 80t - 16t^2 + 3.50 = -16t^2 + 80t + 3.5 Solution of the quadratic equation: x = -0.04 the answer to this is impossible, because our time is negative x = 5.04, which we can round to 5 secondsThe catcher will have 5 seconds before the ball hits the ground.
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