The function is: f ( t ) = - 16 t² + 35 t + 0.25 We can use the derivative: f ` ( t ) = - 32 t + 35 - 32 t + 35 = 0 t = 35 : 32 = 1.09375 s The maximum height is: f ( 1.09375 ) = - 16 · ( 1.09357 )² + 35 · 1.09357 + 0.25 = = - 19.14 + 38.28125 + 0.25 = 19.39 ft ≈ 19.4 ft. Answer: The approximate maximum height of the ball is 19.4 ft.