Question

Plzzz answer this question..​

Answer 1

Answer:

We 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 :

$d = u \cos( \theta) t$

Along the Y axis :

$h = u \sin( \theta) t - \frac{1}{2} g {t}^{2}$

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

$= > h = u \sin( \theta) \{ \dfrac{d}{u \cos( \theta) } \} - \dfrac{1}{2} g { \{ \dfrac{d}{ u\cos( \theta) } \} }^{2}$

$= > h = d \tan( \theta) - \dfrac{g {d}^{2} }{2 {u}^{2} { \cos}^{2} (\theta) }$

$= > \dfrac{g {d}^{2} }{2 {u}^{2} { \cos}^{2} (\theta) } = d \tan( \theta) - h$

$= > 2 {u}^{2} { \cos}^{2} ( \theta) = \dfrac{g {d}^{2} }{ \{d \tan( \theta) - h \}}$

$= > u = \dfrac{d}{ \cos( \theta) } \sqrt{ \dfrac{g}{2 \{d \tan( \theta) - h \}} }$

So final answer :

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