Question

A construction worker tosses a brick from a tall building. The brick's height (in meters above the ground) t tt seconds after being thrown is modeled by h ( t ) = − 5 t 2 + 20 t + 105 h(t)=−5t 2 +20t+105h, left parenthesis, t, right parenthesis, equals, minus, 5, t, squared, plus, 20, t, plus, 105 Suppose we want to know the height of the brick above the ground at its highest point. 1) Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation. h ( t ) = h(t)=h, left parenthesis, t, right parenthesis, equals 2) At its highest point, how far above the ground was the brick?

Answer 1

The quadratic function of the brick's height is rewritten in vertex form to find that the brick reaches its highest point at 2 seconds with a maximum height of 125 meters above the ground.

Step-by-step explanation:

The student's question involves a kinematics problem modeled by a quadratic function, which is essential to find the highest point of the brick's trajectory. The height function given is h(t) = -5t^2 + 20t + 105, which is a downward-facing parabola because the coefficient of t^2 is negative. To find the height of the brick at the highest point, we need to rewrite the function in vertex form to easily identify the vertex, which represents the maximum point in this context.

By completing the square or using the vertex formula h(t) = a(t-h)^2 + k, where (h, k) is the vertex, we get:

h(t) = -5(t-2)^2 + 125

The vertex form reveals that the brick reaches its highest point at t = 2 seconds, and the maximum height is 125 meters above the ground.

Answer 2

h(t)=−5(t−2)^2+125, 125

Step-by-step explanation: