Question

If a stone is to hit at a point which is at a horizontal distance d away and at a height h above the point from where the stone starts, then what is the value of initial speed u if the stone is launched at an angle q ?

• A. $\frac { \mathrm { g } } { \cos \theta }$ $\sqrt{\frac{d}{2(d\tan\theta - h)}}$
• B. $\frac { d } { \cos \theta }$ $\sqrt{\frac{g}{2(d\tan\theta - h)}}$
• C. $\sqrt{\frac{gd^{2}}{h\cos^{2}\theta}}$
• D. $\sqrt{\frac{gd^{2}}{d - h}}$

Answer 1

Answer: (B)

$\frac { d } { \cos \theta }$ $\sqrt{\frac{g}{2(d\tan\theta - h)}}$

Answer 2

Q y = x tan q – $\frac { \mathrm { gx } ^ { 2 } } { 2 \mathrm { u } ^ { 2 } \cos ^ { 2 } \theta }$

Here h = d tan q – $\frac { g d ^ { 2 } } { 2 u ^ { 2 } \cos ^ { 2 } \theta }$

$\frac { g d ^ { 2 } } { 2 u ^ { 2 } \cos ^ { 2 } \theta }$ = (d tan q –h)

or u² = $\frac { g d ^ { 2 } } { 2 \cos ^ { 2 } \theta ( d \tan \theta - h ) }$

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