Question

हाउ मेनी मॉलिक्यूल 79.8 ग्राम ऑफ वाटर?​

Answer 1

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Explanation:

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idkWe shall consider the Projectile motion to be simultaneously occuring two linear motions (one in the X axis and another in Y axis)

Along the X axis :

d = u \cos( \theta) td=ucos(θ)t

Along the Y axis :

h = u \sin( \theta) t - \frac{1}{2} g {t}^{2}h=usin(θ)t−21gt2

Putting value of t from 1st equation into the 2nd equation , we get :

= > h = u \sin( \theta) \{ \dfrac{d}{u \cos( \theta) } \} - \dfrac{1}{2} g { \{ \dfrac{d}{ u\cos( \theta) } \} }^{2}=>h=usin(θ){ucos(θ)d}−21g{ucos(θ)d}2

= > h = d \tan( \theta) - \dfrac{g {d}^{2} }{2 {u}^{2} { \cos}^{2} (\theta) }=>h=dtan(θ)−2u2cos2(θ)gd2

= > \dfrac{g {d}^{2} }{2 {u}^{2} { \cos}^{2} (\theta) } = d \tan( \theta) - h=>2u2cos2(θ)gd2=dtan(θ)−h

= > 2 {u}^{2} { \cos}^{2} ( \theta) = \dfrac{g {d}^{2} }{ \{d \tan( \theta) - h \}}=>2u2cos2(θ)={dtan(θ)−h}gd2

= > u = \dfrac{d}{ \cos( \theta) } \sqrt{ \dfrac{g}{2 \{d \tan( \theta) - h \}} }=>u=cos(θ)d2{dtan(θ)−h}g

So final answer :

\boxed{ \red{ \bold{ \huge{u = \dfrac{d}{ \cos( \theta) } \sqrt{ \dfrac{g}{2 \{d \tan( \theta) - h \}} } }}}}u=cos(θ)d2{dtan(θ)−h}g