Question

if two stones are thrown vertically upwards with same velocity but different projectile angle , their projectile range are same .if one of the projectile angle is π/3 and maximum height y₁ then find the another stone's maximum height


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Answer 1

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h 1 /3

Assuming the angle of projection is = π/3 radians:2π− 3π= 6π radians

⇒ the angle of projection of the other object

let the maximum heights be: h 1

and h 2 : h 2

h 1 = sin 2 ( 6π )sin 2 ( 3π )h 2h 1 =( 21 )

2( 23 ) 2h 2

h 1 = ( 41 )( 43 )= 13h 2 = 3h 1

Explanation:

Hope it will helpful for you thanks

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Answer 2

Answer:

• Given that, both the stones are thrown vertically upwards with the same initial velocity. Let the velocity be 'u' for both stones.

• Their projectile range are same i.e the horizontal distance travelled by them are same.

• One of the projectile angle is π/3. The angle is given in radian. So, first let's convert it in degrees.

$\longrightarrow \theta = \dfrac{\pi}{ 3}$

We know that, 2π = 360° and π = 180°. Substituting the value of π = 180° in above equation we get :

$\longrightarrow \theta = \dfrac{180}{ 3}$

$\longrightarrow \theta = {60}^{ \circ}$

If the range is same for two angles then that two angles are complementary. Therefore angle of projection 'α' for another stone is :

$\longrightarrow \alpha = {90}^{ \circ} - \theta$

$\longrightarrow \alpha = {90}^{ \circ} - {60}^{ \circ}$

$\longrightarrow \alpha = {30}^{ \circ}$

Maximum height of the first stone is h₁. Now, let's find the maximum height of second stone:

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

Comparing their ratios we have:

$:\implies \sf \dfrac{H_1}{H_2} = \dfrac{u^2 \sin^2 \theta}{u^2\sin^2 \alpha}\times \dfrac{2g}{2g}$

$:\implies \sf \dfrac{H_1}{H_2} = \dfrac{\sin^2 \theta}{\sin^2 \alpha}$

$:\implies \sf \dfrac{H_1}{H_2} = \dfrac{\sin^2 {(60}^{ \circ}) }{\sin^2( {30}^{ \circ}) }$

$:\implies \sf \dfrac{H_1}{H_2} = \dfrac{ \bigg(\frac{ \sqrt{3} }{2} \bigg)^{2} }{ \bigg(\frac{1}{2} \bigg)^{2}}$

$:\implies \sf \dfrac{H_1}{H_2} = \dfrac{ \frac{ 3 }{4} }{ \frac{1}{4}}$

$:\implies \sf \dfrac{H_1}{H_2} = \dfrac{3}{4} \times \dfrac{4}{1}$

$:\implies \sf \dfrac{H_1}{H_2} = 3$

$:\implies \underline{ \boxed{ \orange{\bf H_2= \dfrac{H_1}{3} }}}$