Object Height. Suppose an object is thrown straight up from the ground. The height after 1' seconds is given by the formula h(t) = 3t3 + 83t2 + 209 (a) The time in seconds when the object reached the highest point was Note: Round to 4 digits after the decimal point, if rounding is required 0 13.8333 5 O 18.4444 5 O 0 s O 209 s O 41.5 s 0 None of the other answers (b) The height is maximized at the critical point X = a because the second derivative test found O f\"(a) > O O f\"(a)