A ball thrown vertically upwards falls back on the ground after 6… - Vidyadip

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 6. Application of derivatives→6. Application of derivatives — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If A and B are two events such that $\text{P(A)}=\frac{3}{8},\text{P(B)}=\frac{5}{4}.$ and $\text{P}(\text{A}|\text{B})\times\text{P}(\overline{\text{A}}\cap\text{B})$ is equals to.

$\frac{2}{5}$
$\frac{3}{8}$
$\frac{3}{20}$
$\frac{6}{25}$

The volume ( $V$ ) of the casted half cylinder will be

Two or more vectors having the same initial point are:

Coinitial vectors
colinear vectors
equal vectors
Cannot say

If $y ={x^{{x^2}}}$ then $\frac{{dy}}{{dx}}=$

Choose the correct answer from the given four options.
Let f : R → R be given by f(x) = tanx. Then f^-1(1) is:

$\frac{\pi}{4}$
$\{\text{n}\pi+\frac{\pi}{4}:\text{n}\in\text{Z}\}$
$\text{Does not exist.}$
$\text{None of these}.$

If $f(x)=x^3+b x^2+c x+d$ and 0 < $b^2$ < c then in the interval $(-\infty, \infty)$ , f(x) is

If $\vec p$ and $\vec q$ are unequal unit vectors such that $\left( {\vec p - \vec q} \right) . \left( {\left( {2\vec q + \vec p} \right) \times \left( {3\vec p - \vec q} \right)} \right) = \left| {\vec p + \vec q} \right|$ , then angle between $\vec p$ and $\vec q$ will be

$\int_{}^{} {\sqrt {1 + {x^2}} \;dx = } $

The area between the curve $y = {\sin ^2}x,$ $x - $ axis and the ordinates $x = 0$ and $x = \frac{\pi }{2}$ is

Let the solution curve $y = y ( x )$ of the differential equation $\quad \frac{d y}{d x}-\frac{3 x^5 \tan ^{-1}\left(x^3\right)}{\left(1+x^6\right)^{\frac{3}{2}}} y=2 x$ $\exp \frac{x^3-\tan ^{-1} x^3}{\sqrt{(1+x)^6}}$ pass through the origin. Then $y (1)$ is equal to: