Q1. [4] § 2.3 Relationship between Zeroes and Coefficients of a Polynomial § 2.4 Summary
A ball is thrown in the air so that $t$ seconds after it is thrown, its height $h$ metre above its starting point is given by the polynomial $h = 25t - 5t^2$.
Observe the graph of the polynomial and answer the following questions:
A ball is thrown in the air so that $t$ seconds after it is thrown, its height $h$ metre above its starting point is given by the polynomial $h = 25t - 5t^2$. Observe the graph of the polynomial and answer the following questions:
- (i) Write zeroes of the given polynomial. [1]
- (ii) Find the maximum height achieved by ball. [1]
- (iii) After throwing upward, how much time did the ball take to reach to the height of 30 m? OR Find the two different values of $t$ when the height of the ball was 20 m. [2]
Previously asked in CBSE board exam
2024 30/2/1 Q36
Generated by claude-sonnet-4-6 · 2026-06-15 10:34 · grounding stimulus
Model Answer
(i) Zeroes of the polynomial:
Setting $h = 0$: $25t - 5t^2 = 0 \Rightarrow 5t(5 - t) = 0$
∴ $t = 0$ and $t = 5$
The zeroes are 0 and 5.
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(ii) Maximum height:
Maximum occurs at $t = \dfrac{0+5}{2} = 2.5$ s
$h = 25(2.5) - 5(2.5)^2 = 62.5 - 31.25 = \mathbf{31.25 \text{ m}}$
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(iii) Time to reach 30 m:
$25t - 5t^2 = 30$
$5t^2 - 25t + 30 = 0$
$t^2 - 5t + 6 = 0$
$(t-2)(t-3) = 0$
$t = 2$ s or $t = 3$ s
The ball reaches 30 m at $t = 2$ seconds (while going up).
OR
For $h = 20$ m: $25t - 5t^2 = 20$
$5t^2 - 25t + 20 = 0$
$t^2 - 5t + 4 = 0$
$(t-1)(t-4) = 0$
∴ $t = \mathbf{1 \text{ s}}$ and $t = \mathbf{4 \text{ s}}$
Source: Polynomials (Chapter 2), quadratic polynomial application
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Explanation
- (i) Zeroes = values of $t$ where $h = 0$ (parabola cuts t-axis). Always factorise clearly.
- (ii) Maximum of a parabola is at the midpoint of its zeroes, or substitute $t = 2.5$ directly.
- (iii) Both cases reduce to a standard quadratic — factorise neatly. Note: in the main question, both values ($t=2$ and $t=3$) are valid times; the question asks for when going up, so $t=2$ is the expected answer. In the OR, both values must be stated for full marks.
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