Download App

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

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 6. Application of derivatives1 Mark
MCQ
A ball thrown vertically upwards falls back on the ground after  $6$ second. Assuming that the equation of motion is of the form $s = ut - 4.9{t^2}$, where s is in metre and  $t$ is in second, find the velocity at $t = 0$ .......... $m/s$.
  • A
    $0$
  • B
    $1$
  • ✓
    $29.4$
  • D
    None of these
✓

Answer

Correct option: C.
$29.4$
c
(c) Velocity of ball $(v) =\frac{{ds}}{{dt}} = $$u - 9.8\,t$

Here terminal velocity $v = 0$ and $t = 3\,sec$

$u = 9.8(3) = 29.4 \,m/sec$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 6. Application of derivatives→STD 12 - 6. Application of derivatives — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

If $P\left(A^{\prime}\right)=\frac{3}{5}$ and $P(A \cap B)=\frac{1}{5}$ then the value of $P(B / A)$ will be
→
Let a tangent be drawn to the ellipse $\frac{x^{2}}{27}+y^{2}=1$ at $(3 \sqrt{3} \cos \theta, \sin \theta)$ where $\theta \in\left(0, \frac{\pi}{2}\right)$. Then the value of $\theta$ such that the sum of intercepts on axes made by this tangent is minimum is equal to ..... .
→
Let $S=\{1,2,3, \ldots, 10\}$. Suppose $M$ is the set of all the subsets of $S$, then the relation $R=\{(A, B): A \cap B \neq \phi ; A, B \in M\}$ is :
→
Find the principal values of: $\cot ^{-1}(-\sqrt{3})$
→
The equation of motion of a particle is $\text{s} = \text{2t}^2 + \sin\text{2t,}$ where $s$ is in metres and $t$ is in seconds. The velocity of the particle when its acceleration is $2m/\sec^2$, is:
→
If $A$ and $B$ are events such that $P(A)>0$ and $P(B) \neq 1$, then $P\left(A^{\prime} \mid B^{\prime}\right)$ equals
→
The function $\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$ is :
→
If the position vectors of $A $ and $ B$ are $i + 3j - 7k$ and $5i - 2j + 4k,$ then the direction cosine of $\overrightarrow {AB} $ along $y-$ axis is
→
Let $y=y(x)$ be the solution of the differential equation, $x y^{\prime}-y=x^{2}(x \cos x+\sin x), x>0$ If $y (\pi)=\pi,$ then $y ^{\prime \prime}\left(\frac{\pi}{2}\right)+ y \left(\frac{\pi}{2}\right)$ is equal to
→
Let $\text{U}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{V}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ then $\frac{\text{dU}}{\text{dV}}=$
→