Precalculus Math Problem

Question

I'm not sure how to approach this problem and I'd really appreciate any help!

If a ball is thrown directly upward with a velocity of 36 ft/s, its height (in feet) after t seconds is given by y = 36t − 16t2. What is the maximum height attained by the ball? (Round your answer to the nearest whole number.)

Answer 1

The maximum height occurs at the vertex of the height-vs.-time parabola.

 

h(t) = -16t² + 36t = At² + Bt + C

 

The vertex is at 

 

t = -B/(2A) = -36/[2(-16)] sec = 1.125 sec

 

Evaluate h(t) at t = 1.125 sec to get the maximum height.

Answer 2

To solve you can graph this or make a chart of values for time t and height y.  At a time of 1 second, 36-16 = 20'.   1.2 seconds, the ball will be 20.16 feet in height. At 1.3 seconds, subbing, it is already descending to 19.76...try it!