an object thrown directly upward is at a height of h feet after t seconds , where h = - 14 ( t - 3 ) ^ 2 + 140 . at what height , in feet , is the object 2 seconds after it reaches its maximum height ?
"we see that h will be a maximum h = 140 when t - 3 = 0 , that is when t = 3 . at t = 5 , h = - 14 ( 5 - 3 ) ^ 2 + 140 = - 14 ( 4 ) + 140 = 84 the answer is d ."
add(multiply(negate(14), power(subtract(add(3, 2), 3), 2)), 140)