You hit a ball from 4 ft above the ground with an initial vertical velocity of v ft/s. The function h = −16t2 + vt + 4 models the height h in feet of the ball at time t in seconds. Explain how to find the initial velocity for which the ball reaches a maximum height of 40 ft. A. The maximum or minimum of a quadratic function occurs twice. If the ball reaches a maximum height of 40 ft, then 40 = −16t2 + vt + 4 or 0 = −16t2 + vt − 36 has two real solutions and the discriminant is greater than zero. Solve v2 − 4(−16)(−36) = 0 to get v = ±48. The positive initial velocity is 48 ft/s
B. The maximum or minimum of a quadratic function doesn't occur. If the ball reaches a maximum height of 40 ft, then 40 = −16t2 + vt + 4 or 0 = −16t2 + vt − 36 has no real solution and the discriminant is less than zero.
C. The maximum or minimum of a quadratic function occurs only once. If the ball reaches a maximum height of 40 ft, then 40 = −16t2 + vt + 4 or 0 = −16t2 + vt − 36, has one real solution and the discriminant is zero. Solve v2 − 4(−16)(−36) = 0 to get v = ±48. The positive initial velocity is 48 ft/s.