Projectile model and the 25% claim
A cricket coach models the height of a ball after a vertical hit using h(t) = −5t² + 20t + 1, where h is metres above the ground and t is seconds after the hit. The model assumes no air resistance and takes g = 10 m/s². A second, flatter shot is modelled by g(t) = −4t² + 16t + 1. The coach claims that if the initial upward speed in the first model is raised from 20 m/s to 25 m/s while the launch height and g stay the same, the ball's hang time will increase by exactly 25%. He also wants to know when a fielder standing at the boundary can catch the first ball on its way down at a height of 1.5 m. The coach is unsure whether the '25%' claim is an exact consequence of the model or only a rough estimate.
1. For h(t) = −5t² + 20t + 1, find the maximum height and the hang time until h = 0.
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Complete the square: h = −5(t² − 4t) + 1 = −5[(t − 2)² − 4] + 1 = 21 − 5(t − 2)². So the maximum height is 21 m at t = 2 s.\n\nSet h = 0: 5t² − 20t − 1 = 0. Thus t = [20 ± √(400 + 20)] / 10 = [20 ± √420] / 10. The positive root is (20 + 20.4939) / 10 = 4.049 s. The negative root is rejected because the model's domain is t ≥ 0. Hang time is approximately 4.05 s.
2. Evaluate the coach's exact-25% claim for an initial speed of 25 m/s. Support the answer with calculation.
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New model: h₂(t) = −5t² + 25t + 1. Setting h₂ = 0 gives 5t² − 25t − 1 = 0, so t = [25 + √(625 + 20)] / 10 = [25 + √645] / 10. Since √645 ≈ 25.3969, t ≈ 5.0397 s.\n\nThe ratio is 5.0397 / 4.0494 ≈ 1.2446, an increase of about 24.46%, not exactly 25%. The claim is a close approximation, not an identity. The reason is the constant launch height: if the launch height were 0, hang time would be 2v/g, so raising v from 20 to 25 would increase hang time by exactly 25%. The +1 m term breaks that proportionality.
3. At what time does the original ball pass height 1.5 m on the way down? Why are there two solutions?
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Set −5t² + 20t + 1 = 1.5, so −5t² + 20t − 0.5 = 0, or 5t² − 20t + 0.5 = 0. Then t = [20 ± √(400 − 10)] / 10 = [20 ± √390] / 10. Since √390 ≈ 19.7484, the roots are t ≈ 0.0252 s and t ≈ 3.9748 s.\n\nThe first root is on the way up, almost immediately after launch; the second is on the way down. The fielder's catch on the way down is at approximately 3.97 s. Two roots appear because the ball crosses the same height twice in its upward and downward motion.