If a ball is thrown directly upward with a velocity of 44 feet per second, its height (in feet) after t seconds is given by y=44t-16t2 . What is the maximum height attained by the ball?

Question

If a ball is thrown directly upward with a velocity of 44 feet per second, its height (in feet) after t seconds is given by y=44t-16t² . What is the maximum height attained by the ball?

answer in feet

Answer 1

If we put this in quadratic standard form, we get y = -16t²+44t . Recall that standard form is y = at²+bt+c.

Then we can use this expression to find the x-coordinate of the parabola's vertex: (-b/2a).

In the given equation, b = 44 and a = -16. So, the x-coordinate of the vertex is (-44/2(-16)). Simplify to (11/8).

Now we have the x-coordinate of the vertex (the highest point on the parabola). We can plug that into the equation to get the y-coordinate of the vertex, which will be the maximum height of the ball.

y = -16(11/8)² + 44(11/8)

y = -16(121/64) + 44(11/8)

y = (-121/4) + (121/2)

y = 121/4

y = 30.25

The maximum height of the ball is 30.25 feet.