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 - 5t². Observe the graph of the polynomial and answer the following questions: (i) Write zeroes of the given polynomial. (ii) Find the maximum height achieved by ball. (iii) (a) After throwing upward, how much time did the ball take to reach to the height of 30 m? (iii) (b) Find the two different values of t when the height of the ball was 20 m.

MathsCBSE Class 10 · Standard · Board exam 2024, set 30/2/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2024: question 36

Set in the CBSE Class 10 board exam of 2024, paper 30/2/1, section E, question 36, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q36 · 4 marks · Case-basedCompetency-basedPolynomials

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 $h = 25 t - 5 t^{2}$ . Observe the graph of the polynomial and answer the following questions:

(i)
Write zeroes of the given polynomial.
1 mark
(ii)
Find the maximum height achieved by ball.
1 mark
(iii)
(a) After throwing upward, how much time did the ball take to reach to the height of $30$ m?
2 marks
OR
(iii)
(b) Find the two different values of $t$ when the height of the ball was $20$ m.
2 marks

Figure in the paper: Graph of h(t) on vertical axis against t on horizontal axis. The curve is a downward-opening parabola starting at the origin 0, rising to a maximum point, and returning to the t-axis. A dashed vertical line marks the peak, labelled (5/2, 0) on the t-axis below it. The curve meets the t-axis again at (5, 0). See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 30/2/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Zeroes are the values of $t$ where the graph cuts the $t$ -axis.

Hint 2

The highest point of the parabola lies exactly halfway between the two zeroes. Put that value of $t$ into $h$ .

Hint 3

For (iii), set $25 t - 5 t^{2}$ equal to the given height, divide by $- 5$ , and factorise the quadratic. Think about which root belongs to the upward journey.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) The graph of $h = 25 t - 5 t^{2} = 5 t (5 - t)$ meets the $t$ -axis at $t = 0$ and $t = 5$ . So the zeroes of the polynomial are $0$ and $5$ .
1 mark
(ii) The maximum occurs at the vertex of the parabola, at $t = \frac{5}{2}$ (the dashed line in the graph). Substituting: $h = 25 \times \frac{5}{2} - 5 \times (\frac{5}{2})^{2}$
½ mark
$h = \frac{125}{2} - \frac{125}{4} = \frac{125}{4} = 31.25$ m. So the maximum height is $31.25$ m.
½ mark
(iii)(a) Put $h = 30$ : $25 t - 5 t^{2} = 30$
½ mark
Dividing by $- 5$ and rearranging: $t^{2} - 5 t + 6 = 0$
½ mark
Factorising: $(t - 2) (t - 3) = 0$ , so $t = 2$ or $t = 3$ .
½ mark
The ball goes up first and reaches 30 m at $t = 2$ s. It is at 30 m again at $t = 3$ s while coming down. So the ball takes 2 seconds to reach 30 m on the way up.
½ mark
(iii)(b) Put $h = 20$ : $25 t - 5 t^{2} = 20$
½ mark
Dividing by $- 5$ and rearranging: $t^{2} - 5 t + 4 = 0$
½ mark
Factorising: $(t - 1) (t - 4) = 0$
½ mark
$t = 1$ s (going up) and $t = 4$ s (coming down).
½ mark

Answer: (i) 0 and 5; (ii) 125/4 m = 31.25 m; (iii)(a) 2 seconds; (iii)(b) t = 1 s and t = 4 s

Where the marks are

(i) Writes the zeroes as 0 and 5 (1 mark)
(ii) Substitutes $t = \frac{5}{2}$ into $h$ (½ mark)
(ii) Gets the maximum height $\frac{125}{4}$ m or 31.25 m (½ mark)
(iii)(a) Forms $25 t - 5 t^{2} = 30$ (½ mark)
(iii)(a) Reduces it to $t^{2} - 5 t + 6 = 0$ (½ mark)
(iii)(a) Factorises as $(t - 2) (t - 3) = 0$ (½ mark)
(iii)(a) Rejects $t = 3$ for the upward journey and states $t = 2$ s (½ mark)
(iii)(b) Forms $25 t - 5 t^{2} = 20$ (½ mark)
(iii)(b) Reduces it to $t^{2} - 5 t + 4 = 0$ (½ mark)
(iii)(b) Factorises as $(t - 1) (t - 4) = 0$ (½ mark)
(iii)(b) States $t = 1$ s and $t = 4$ s (½ mark)

Revise the chapter

Chapter: Polynomials (notes, formulas, where marks are lost and every past question)
Topic: Forming a Quadratic Polynomial from Its Zeros

Practise this chapterBoard-style questions from Polynomials, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 30/2/1 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard 2024: question 36

Hints

Step-by-step solution

Revise the chapter

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