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Mathematics · 2024 · 4 marks
CBSE 2024 · Region 2 · Set 1 · Q36
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…
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 $\displaystyle \mathrm{h}=25 \mathrm{t}-5 \mathrm{t}^{2}$.
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 $\displaystyle 30$ m ?
OR
(b)
Find the two different values of t when the height of the ball was $\displaystyle 20$ m.
Marking-scheme solution
(i)
Zeroes of the polynomial are $\displaystyle 0$ and $\displaystyle 5$
(ii)
Maximum height achieved by ball
\[\begin{array}{l}
=25 \times \frac{5}{2}-5 \times\left(\frac{5}{2}\right)^{2} \\
=\frac{125}{4} \text { or } 31.25 \mathrm{~m}
\end{array}
\]
(iii)
$\displaystyle -5 \mathrm{t}^{2}+25 \mathrm{t}=30$
\[\begin{aligned}
\Rightarrow & t^{2}-5 t+6=0 \\
\Rightarrow & (t-2)(t-3)=0 \\
& t \neq 3, t=2
\end{aligned}
\]
$\displaystyle -5 \mathrm{t}^{2}+25 \mathrm{t}=20$
\[\begin{array}{l}
\Rightarrow \mathrm{t}^{2}-5 \mathrm{t}+4=0 \\
\Rightarrow(\mathrm{t}-4)(\mathrm{t}-1)=0 \\
\Rightarrow \mathrm{t}=4,1
\end{array}
\]
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CBSE Class 10 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.