Question

A ball is thrown directly upward from an initial height of 200 feet with an initial velocity of 96 feet per second. The height of the ball thrown is given by the function n(t) =-167 + 96t + 200. How long until the ball reaches the ground? How long until the ball is 100 feet above the ground?

Answer 1

To find the times when the ball reaches the ground and is at 100 feet, the quadratic formula must be applied to the height function which represents the ball's trajectory.

Step-by-step explanation:

The student's question involves finding the time when a ball thrown upwards reaches the ground and when it is at a certain height in its trajectory, specifically 100 feet above the ground. The function n(t) = -16t2 + 96t + 200 (corrected for standard gravity in feet per second squared) describes the height of the ball over time, where t is the time in seconds.

To find when the ball reaches the ground, we set n(t) equal to 0 and solve for t using the quadratic formula t = (-b ± √(b2 - 4ac)) / (2a). This represents free fall of a ball.

Similarly, to find when the ball is at 100 feet, we set n(t) equal to 100 and again solve for t using the quadratic formula.