A stone is projected from a point on the ground in a direction to hit – InterviewSolution

1.

A stone is projected from a point on the ground in a direction to hit a bird on the top of a telegraph post of height h, it attains a maximum height 2h above the ground otherwise. If at the in.stant of projection the bird were to fly away horizontally with a uniform speed, find the ratio of the horizontal velocities of the bird and the stone if the stone still hits the bird

Answer»

$\bf\huge{ANSWER}$

ASSUME THAT :

Let's consider θ be the angle of projection and u be the velocity of projection.
According to the situation shown in figure it is given that maximum height of projectile is 2 h ,we have .

$\bf{ = > u \: sin \: θ = \sqrt{4gh} }$

• => If time taken by the projectile to reach points A and B are t 1 and t 2 respectively then ROOTS of equations are t 1 and t 2 .

Here now refer the attachment :

Solving :

$\bf{= > t = \frac{u \: sin \: θ}{g} ± \frac{ \sqrt{u \: sin \:θ - 2gh} }{g} } \\$

USING :

$\bf {= > t = \sqrt{ \frac{4h}{g} } ± \sqrt{ \frac{2h}{g} } } \\$

Here we have :

$\bf{ = > t \: 1 = \sqrt{ \frac{4h}{g} } - \sqrt{ \frac{2h}{g} } } \\$

$\bf{and}$

$\bf{ = > t \: 2 = \sqrt{ \frac{4h}{g} } - \sqrt{ \frac{2h}{g} } } \\$

Here :

Now distance AB can be WRITTEN as

$\bf{ = > v \: t \: 2 = u \: cos \: θ(t \: 2 - t \: 1)} \\$

Ratio of horizontal velocity :

$\bf{ = > \frac{v}{u \: cos \: θ} = \frac{t \: 2 - t \: 1}{t \: 2} } \\$

$\bf \large{ = > \frac{2}{ \sqrt{2} + 1 } } \\$

$\bf \boxed{ = > \frac{2}{ \sqrt{2} + 1 } }$

#Answer with quality
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NOTE:

Please refer the 2 nd attachment to see second method to solve this question
Kindly refer all THREE attachment.

________________

Here is User

ALEXANDER ANSWER

[tex]\bf\huge\orange{ANSWER}[/tex]

$\bf\huge\pink{Here}$ $\bf{ = >2h = \frac{Vo²si {n}^{2} θ} {2g} } \\$

$\bf{ = >Vo = \frac{2 \sqrt{gh} }{si {n}^{2}θ } } \\$

$\bf\huge\pink{Also}$

$\bf{ = > y = Vo \: sin \: θ \: t - \frac{1}{2} \: g {t}^{2} }\\$

$\bf{ = > h = \frac{2 \sqrt{gh} }{sin \: θ}.sin \: θ \: t - \frac{1}{2} {gt}^{2} } \\$

$\bf{ = > {gt}^{2} - 4 \sqrt{gh}t + 2h = 0} \\$

$\bf{ = > t = \frac{4 \sqrt{gh}±2 \sqrt{2gh} }{2g} } \\$

$\bf {= > t \: 1 = \sqrt{ \frac{h}{g} } \: (2 - 2 \sqrt{2} )} \\$

$\bf {= > t \: 2 = \sqrt{ \frac{h}{g} } \: (2 + 2 \sqrt{2} )} \\$

$\bf\huge\red{Therefore}$ $\bf{ = > Vt²=Va \: Cos \: θ (t2 -t1 )}$

$\bf{ = > \frac{V}{Va \: Cos \: θ } = \frac{t \: 2 - \: t \: 1}{t \: 2} } \\$ $\bf { = > \frac{2}{ \sqrt{2} + 1 } } \\$

$\bf\huge\blue{Ratio}$ $\sf \huge{ = > \frac{2}{ \sqrt{2} + 1} }$

Note:

Kindly refer the attachment to see full answer solve by USER ALEXANDER

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