Question

A ball is thrown vertically upward from a height of

40 m and hits the ground with a speed that is three

times its initial speed. What is the time taken (in

sec) for the fall?​

Answer 1

Answer:

u=x

v=-3x

s=-40m

a=-g=-10m/S2

t=?

using the equation V2-U2=2as

9x2-x2=2*10*40

8x2=800

X=10

Now V=U+at

-3x=x-10t

10t= 40

t =4 sec

Answer 2

Answer:

$\sf{The \ the \ time \ taken \ for \ fall}$

$\sf{is \ 3.03 \ s}$

Given:

$\sf{v_{2}=3\times \ u_{1}}$

$\sf{Ball \ is \ thrown \ upward \ from \ height \ of \ 40 \ m}$

To find:

• $\sf{Time \ taken \ for \ the \ fall.}$

Solution:

$\sf{Let \ initial \ velocity \ of \ ball \ while \ throwing}$

$\sf{be \ u_{1}}$

$\sf{For \ upward \ direction \ final \ velocity \ v_{1}=0}$

$\sf{According \ to \ the \ third \ equation \ of \ motion}$

$\sf{v_{1}^{2}=u_{1}^{2}+2as}$

$\sf{\therefore{s=\dfrac{-u_{1}^{2}}{2(-9.8)}}}$

$\sf{\therefore{s=\dfrac{u_{1}^{2}}{19.6}}}$

$\sf{\therefore{Total \ height=40+\dfrac{u_{1}^{2}}{19.6}}}$

$\sf{For \ downward \ motion}$

$\sf{v_{2}=3u_{1}}$

$\sf{u_{2}=0}$

$\sf{According \ to \ third \ equation \ of \ motion}$

$\sf{(3u_{1})^{2}=0+2(9.8)(40+\dfrac{u_{1}^{2}}{19.6})}$

$\sf{\therefore{9u_{1}^{2}=19.6(40+\dfrac{u_{1}^{2}}{19.6})}}$

$\sf{\therefore{9u_{1}^{2}=784+u_{1}^{2}}}$

$\sf{\therefore{8u_{1}^{2}=784}}$

$\sf{\therefore{u_{1}^{2}=\dfrac{784}{8}}}$

$\sf{\therefore{u_{1}^{2}=98}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{\leadsto{u_{1}=9.9 \ m \ s^{-1}}}$

$\sf{But, \ v_{2}=3u_{1}}$

$\sf{\therefore{v_{2}=3\times9.9}}$

$\sf{\therefore{v_{2}=29.7 \ m \ s^{-1}}}$

$\sf{According \ to \ the \ first \ equation \ of \ motion}$

$\sf{v_{2}=u_{2}+at}$

$\sf{\therefore{29.7=0+(9.8)t}}$

$\sf{\therefore{t=\dfrac{29.7}{9.8}}}$

$\sf{\therefore{t=3.03 \ s \ (approx)}}$

$\sf\purple{\tt{\therefore{The \ the \ time \ taken \ for \ fall}}}$

$\sf\purple{\tt{is \ 3.03 \ s}}$

__________________________

$\sf\blue{Extra \ information}$

$\sf{Equations \ of \ motion}$

$\sf{1. \ v=u+at}$

$\sf{2. \ s=ut+\dfrac{1}{2}\times \ at^{2}}$

$\sf{3. \ v^{2}=u^{2}+2as}$