Question

Two particles are projected simultaneously from the top and the bottom of a tower at angles α and β with horizontal. If both the particles strike on the ground ‘d’ distance away from the foot of the tower at the same instant of time then the height of the tower is:

Answer 1

Given:

Two particles are projected simultaneously from the top and the bottom of a tower at angles α and β with horizontal. The objects strike on the ground ‘d’ distance

To find:

Height of the tower

Calculation:

Let height of tower be h ;

For the object from top of tower:

$\therefore \: h = ( u1)_{y}t - \dfrac{1}{2} g {t}^{2}$

$= > \: h = \{(u1) \sin( \alpha ) \} t - \dfrac{1}{2} g {t}^{2}$

For the object at the bottom of the tower:

$\therefore \: d = (u2)_{x}t$

$= > d = (u2) \cos( \beta ) t$

Putting value of t in eq.(1) :

$= > \: h = \{(u1) \sin( \alpha ) \} t - \dfrac{1}{2} g {t}^{2}$

$= > \: h = \{(u1) \sin( \alpha ) \} \times \dfrac{d}{(u2) \cos( \beta ) } - \dfrac{1}{2} g { \bigg \{ \dfrac{d}{(u2) \cos( \beta ) } \bigg \}}^{2}$

$= > \: h = \dfrac{(u1) \sin( \alpha ) d}{(u2) \cos( \beta ) } - \dfrac{g {d}^{2} }{2 {(u2)}^{2} { \cos}^{2} ( \beta ) }$

So, final answer is:

$\boxed{ \bold{ h = \dfrac{(u1) \sin( \alpha ) d}{(u2) \cos( \beta ) } - \dfrac{g {d}^{2} }{2 {(u2)}^{2} { \cos}^{2} ( \beta ) } }}$