Task 2: Quadratic Equations and Inequalities prob Solve the following problem: A ball is thrown upward with an initial velocity of 20 m/s from the ground. Its height h(t) after t seconds is given by h(t) = -5t2 + 20t. a) At what time will the ball reach its maximum height? b) What is the maximum height? c) For what values of t will the ball be at least 15 meters high? Represent this as a quadratic inequality and solve.
Answer (1)
a) To find the time at which the ball reaches its maximum height, we need to find the vertex of the quadratic equation h(t) = -5t^2 + 20t. The t-coordinate of the vertex is given by t = -\frac{b}{2a}, where a = -5 and b = 20.t = -\frac{20}{2(-5)} = -\frac{20}{-10} = 2So, the ball will reach its maximum height at 2 seconds.b) To find the maximum height, we substitute the value of t we found in part (a) into the equation h(t) = -5t^2 + 20t.h(2) = -5(2)^2 + 20(2) = -5(4) + 40 = -20 + 40 = 20Therefore, the maximum height is 20 meters. c) We want to find the values of t for which h(t) \ge 15. So, we set up the inequality:-5t^2 + 20t \ge 15Rearrange the inequality:-5t^2 + 20t - 15 \ge 0Divide by -5 (and reverse the inequality sign):t^2 - 4t + 3 \le 0Factor the quadratic expression:(t - 1)(t - 3) \le 0To find the intervals where the inequality holds, we analyze the sign of (t - 1)(t - 3). The critical points are t = 1 and t = 3. - When t < 1, both (t - 1) and (t - 3) are negative, so their product is positive.- When 1 < t < 3, (t - 1) is positive and (t - 3) is negative, so their product is negative.- When t > 3, both (t - 1) and (t - 3) are positive, so their product is positive.Since we want (t - 1)(t - 3) \le 0, the solution is the interval where the product is negative or zero.Thus, the ball will be at least 15 meters high for 1 \le t \le 3.
