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Object Height. Succose an object thrown straight up from the ground. The beght aftert seconds given by the formula hots-2201²195 (a) The time in seconds when the object reached the highest point was Note Round 4 after the decal point, founding required 0.175 omi 23.3393 Co CONSI One of the (0) The height is maxed at the entical point a because the second devet und Oruwas negative to the left of a ant petive to the righ Orava OFWYD 02-0 Ortal was postive to the left ofa and negative to the right 183-0
Object Height. Suppose an object is thrown straight up from the ground. The height after t seconds is given by the formula h(t) = -2t³ +70t² + 195 (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 17.5 s 195 s O23.3333 s Oos 52.5 s None of the other answers (b) The height is maximized at the critical point x = a because the second derivative test found Or(a) was negative to the left of x = a and positive to the right Of" (a) <0 Of" (a) > 0 Or(a)=0 Or(a) was positive to the left of xa and negative to the right Or" (a) = 0 Submit Answer