Advanced MathHardTime interval a projectile stays at or above its launch height
Advanced Math practice question
A ball is thrown upward from a platform, and its height in metres t seconds after the throw is h(t) = −5t² + 20t + 25. For how many seconds is the ball at or above the height of the platform?
Answer: 4
The platform height is h(0) = 25. Solve h(t) ≥ 25: −5t² + 20t ≥ 0 → 5t(4 − t) ≥ 0 → 0 ≤ t ≤ 4. The ball is at or above the platform for 4 seconds.
Common wrong answers
- If you answered 2
- 2 is the time to reach the top; the ball then takes another 2 seconds to come back down to platform height.
- If you answered 5
- 5 is when the ball hits the ground (h = 0), which is below the platform.
- If you answered 25
- 25 is the platform height, not a time.
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