A ball is thrown upward with speed (20\text{ m/s}). Its height is (h=20t-5t^2). At what times will it be at height (15\text{ m})?
Answer and explanation
Correct answer: (1\text{ s}) and (3\text{ s})
Set the given height equal to 15: \(20t-5t^2=15\u0005. Rearranging and dividing by \(-5\u0005 gives \(t^2-4t+3=0\u0005. Factoring produces \((t-1)(t-3)=0\u0005, so the possible times are \(t=1\u0005 s and \(t=3\u0005 s. Both are physically meaningful because the ball reaches the same height while rising and later while falling.
Therefore option A is correct. Substitution confirms that at either time the height is 15 m: at 1 s, \(20-5=15\u0005, and at 3 s, \(60-45=15\u0005. A single time would miss the two stages of the motion.
Frequently asked questions
What is the correct answer to this question?
(1\text{ s}) and (3\text{ s})
Why is this the correct answer?
Set the given height equal to 15: \(20t-5t^2=15\u0005. Rearranging and dividing by \(-5\u0005 gives \(t^2-4t+3=0\u0005. Factoring produces \((t-1)(t-3)=0\u0005, so the possible times are \(t=1\u0005 s and \(t=3\u0005 s. Both are physically meaningful because the ball reaches the same height while rising and later while falling.
Therefore option A is correct. Substitution confirms that at either time the height is 15 m: at 1 s, \(20-5=15\u0005, and at 3 s, \(60-45=15\u0005. A single time would miss the two stages of the motion.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Word Problems and Applications.
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