Word problem involving maximum

Question

A ball is thrown vertically upward. After t seconds, its height (in feet) is given by the function h(t)=72t-16t^2. After how long will it reach its maximum height.

Answer 1

I see there have already been some responses for how to get the solution algebraically, I will cover how to get it on the TI-84.

1. Enter the equation into y1 by pressing the y=button
2. Press second function - Trace (For Calc menu)
3. Select #4 Max
4. Scroll to the left side of the maximum point, press Enter
5. Scroll to the right side of the maximum point, press Enter
6. Press Enter a 3rd time

The x value is 2.25 seconds

Answer 2

Hello Courtney,

One way to solve this problem is to use the formula x = -b/2a for y = ax^2 + bx + c. This gives you the x-value (coordinate) for the vertex of the parabola which is either the maximum (if it opens down) or the minimum (if it opens up). In your question t is x and h is y. See if you can use the formula to find the answer for the t value at the maximum height since the parabola opens down.

Tom S.

Answer 3

In an equation in the form f(x) = ax² + bx + c the maximum or minimum occurs at -b / 2a

h(t) = -16t² + 72t

Maximum at t = -72 / 2(-16)

Answer 4

The maximum height will be at the vertex . The x coordinate of the vertex is at -b//2a which in this problem is -72/2(-16)= 72/32 = 2.25secinds