Question

The height, h, in feet, of a baseball tossed straight up into the air is modeled by the equation h(t) = -16t^2 + 5t, where t is the number of seconds after the ball was tossed. What was the maximum height of the ball?

Answer 1

The maximum height of the ball is 25/64 feet.

Step-by-step explanation:

The maximum height of the ball can be found by determining the vertex of the quadratic equation. In the equation h(t) = -16t^2 + 5t, the coefficient of t^2 is -16. The vertex of a quadratic equation of the form h(t) = at^2 + bt + c is given by the x-coordinate of the vertex, which is -b/2a. In this case, a = -16 and b = 5, so the x-coordinate of the vertex is -5/(-32), which simplifies to t = 5/32. To find the maximum height, substitute this value of t into the equation h(t). h(5/32) = -16(5/32)^2 + 5(5/32) = -400/1024 + 25/32 = -25/64 + 25/32 = 25/64. Therefore, the maximum height of the ball is 25/64 feet.