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−16t^2. 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

Helo, Axl,

We can solve this by either of two approaches. I like the derivative approach, so I'll start there.

1) Take the first derivative of y=44t-16t². The first derivative tells the slope of this parabolic equation at any time t. If you plot the parabola (y is the height in feet and x is the time, in seconds), there will be an apex that indicates the ball has stopped rising and is starting to fall. The slope at this apex is zero.

y=44t-16t²

y'=44-32t

t=1.375 sec

plug this time into the original equation to find the height when the ball reaches the top.

y = 44*(1.375)-16*(1.375)²

y= 30.25 feet

2) We can also graph the equation and find the coordinates of the apex, which I find to be (1.375, 30.25)

Bob