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Q.Case Study - 3 : A volleyball player serves the ball which takes a parabolic path given by the equation h(t)=−72t2+132t+1h(t) = -\frac{7}{2}t^2 + \frac{13}{2}t + 1, where h(t)h(t) is the height of ball at any time tt (in seconds), (t≥0)(t \geq 0). Based on the above information, answer the following questions :

(i) Is h(t)h(t) a continuous function ? Justify. [2 marks]
(ii) Find the time at which the height of the ball is maximum. [2 marks]
CBSECBSE Class XII Board 2023Subjective· 4mImportance★★★★★
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A polynomial is continuous everywhere; the maximum height occurs at the vertex of the parabola, found by t=−b2at = -\frac{b}{2a}, giving t=1314t = \frac{13}{14} seconds.


The trajectory of the volleyball is modeled by a quadratic function. Before we find when the ball reaches its peak, we need to understand what continuity means and why polynomials behave so nicely.

(i) Continuity of h(t)h(t)

A function is continuous at a point if it has no breaks, jumps, or holes there. More formally, ff is continuous at cc if lim⁡x→cf(x)=f(c)\lim_{x \to c} f(x) = f(c).

The function h(t)=−72t2+132t+1h(t) = -\frac{7}{2}t^2 + \frac{13}{2}t + 1 is a polynomial. Polynomials are built from powers of tt and constants using only addition, subtraction, and multiplication—operations that preserve continuity. Since each term tnt^n is continuous for all real tt, and sums and scalar multiples of continuous functions remain continuous, every polynomial is continuous on its entire domain.

Yes, h(t)h(t) is a continuous function for all t∈Rt \in \mathbb{R}, and in particular for t≥0t \geq 0. There are no breaks or jumps in the ball's trajectory.

Note

Polynomials, rational functions (where the denominator is non-zero), exponentials, logarithms, and trigonometric functions are all continuous on their natural domains. Continuity is the rule, not the exception, for elementary functions.


(ii) Time at which height is maximum

The parabola h(t)=−72t2+132t+1h(t) = -\frac{7}{2}t^2 + \frac{13}{2}t + 1 opens downward (since the coefficient of t2t^2 is negative), so it has a maximum at its vertex.

For a quadratic at2+bt+cat^2 + bt + c, the vertex occurs at t=−b2at = -\frac{b}{2a}. This comes from completing the square or from calculus: the derivative h′(t)=2at+bh'(t) = 2at + b vanishes when t=−b2at = -\frac{b}{2a}. …

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