Calculus or physics question.

When a ball is thrown vertically upward
into the air with a velocity of 53 ft/sec its
height, y(t), in feet after t seconds is given by
y(t) = 53t − 16t^2

Find the average velocity of the ball over
the interval from 3 to 3 + h seconds, h cannot equal 0.

Average velocity = [y(3 + h) - y (h)]/ h

[53(3 + h) - 16(3 + h)^2 - 53(3) + 16(3)^2)]/h

[459 + 53h - 144 - 96h - 16h^2 - 459 + 144]/h = [-43h - 16h^2]/h = -43 - 16h = -(43 + 16h) ft/sec