If a ball is thrown directly upward with a velocity of 48 ft/s, its height (in feet) after t seconds is given
by $y = 48 t - 16 t ^ { 2 }$ . What is the maximum height attained by the ball?

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The Answer of If a ball is thrown directly upward with a velocity of 48 ft/s, its height (in feet) after t seconds is given
by $y = 48 t - 16 t ^ { 2 }$ . What is the maximum height attained by the ball?

If a Ball Is Thrown Directly Upward with a Velocity y=48t−16t2y = 48 t - 16 t ^ { 2 }y=48t−16t2

If a Ball Is Thrown Directly Upward with a Velocity $y = 48 t - 16 t ^ { 2 }$