O-Level Quadratic functions: worked solution

Question

For the projectile with h = 42 - (1/250)(t - 100)^2, find the length of time during which its height stays at 32 m or more above the ground.

Step 1: turn the wording into an inequality. "Height stays at 32 m or more" means h ≥ 32. 42 − (1/250)(t − 100)² ≥ 32 Step 2: isolate the squared term. Subtract 42 from both sides. −(1/250)(t − 100)² ≥ −10 Step 3: multiply both sides by −250 to clear the fraction and the minus sign. Multiplying an inequality by a negative number reverses the direction of the sign. (t − 100)² ≤ 2500 Step 4: take the square root. A square that is less than or equal to a constant gives a two-sided band, not a single root, because both a positive and a negative value can square to 2500. −50 ≤ t − 100 ≤ 50 Step 5: add 100 to all three parts. 50 ≤ t ≤ 150 So the projectile is at least 32 m up from t = 50 s until t = 150 s. Step 6: the length of time is the width of that interval, not its end value. 150 − 50 = 100 The height stays at 32 m or more for 100 s.