Question

the range of two projectiles launched from ground with the same initial velocity u and angles of projection theta 1 and theta 2 are equal .the ratio of maximum heights reaches by them is h1/h2 =1/3 .the range of projectiles R is equal to

a. u²/2g
b. 2u²/ g
c. u² /4g
d. √3u²/2g
​

Answer 1

Given:

The range of two projectiles launched from ground with the same initial velocity u and angles of projection theta 1 and theta 2 are equal .the ratio of maximum heights reaches by them is h1/h2 =1/3.

To find:

Value of range ?

Calculation:

Always remember that :

• Two projectiles when thrown at complementary angle of projections with the same speed have same value of range.

So, if one angle of projection is $\theta$ , then the other angle of projection will be (90°-$\theta)$.

Since h1 : h2 = 1 : 3 , let the heights be x and 3x.

For 1st Projectile:

$\therefore \: h1 = \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g}$

$\implies\: x= \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g}$

For 2nd Projectile:

$\therefore \: h2= \dfrac{ {u}^{2} { \sin}^{2}( {90}^{ \circ} - \theta) }{2g}$

$\implies \: h2= \dfrac{ {u}^{2} { \cos}^{2}( \theta) }{2g}$

$\implies \:3x= \dfrac{ {u}^{2} { \cos}^{2}( \theta) }{2g}$

Dividing the equations:

$\therefore \: \dfrac{ { \sin}^{2}( \theta) }{ { \cos}^{2} ( \theta)} = 3$

$\implies \: { \tan}^{2} ( \theta) = 3$

$\implies \: \tan( \theta) = \sqrt{3}$

$\implies \: \theta= {60}^{ \circ}$

Now , let range be R:

$\therefore \: R = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$\implies \: R = \dfrac{ {u}^{2} \sin(2 \times {60}^{ \circ} ) }{g}$

$\implies \: R = \dfrac{ {u}^{2} \sin( {120}^{ \circ} ) }{g}$

$\implies \: R = \dfrac{ {u}^{2} \sin( {180}^{ \circ} - {60}^{ \circ} ) }{g}$

$\implies \: R = \dfrac{ {u}^{2} \sin( {60}^{ \circ} ) }{g}$

$\implies \: R = \dfrac{ {u}^{2} \times \frac{ \sqrt{3} }{2} }{g}$

$\implies \: R = \dfrac{ \sqrt{3} {u}^{2} }{2g}$

So, final answer is:

$\boxed{ \bold{ \: R = \dfrac{ \sqrt{3} {u}^{2} }{2g} }}$

Answer 2

Answer:

√3u2/2g your answer this.. .