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STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 6. Application of derivatives1 Mark
MCQ
The maximum height is reached in $5$ seconds by a stone thrown vertically upwards and moving under the equation  $10s = 10ut -49{t^2}$, where $ s$ is in metre and $t$ is in second. The value of $u$ is ........ $m/\sec $
  • A
    $4.9$
  • ✓
    $49$
  • C
    $98$
  • D
    None of these
✓

Answer

Correct option: B.
$49$
b
(b) Given equation is $\frac{{dy}}{{dt}} = - \frac{{16}}{6} = - \frac{8}{3}cm/\sec .$ or $s = ut - 4.9{t^2}$

==> $\frac{{ds}}{{dt}} = u - 9.8t = v$

When stone reached the maximum height, then $v = 0$

==> $u - 9.8t = 0 \Rightarrow u = 9.8\,t$

But time $t = 5\,sec$

So the value of $u = 9.8 \times 5 = 49.0\,m/sec$

Hence initial velocity $ = 49\,m/sec.$

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