Question

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 1

$\bf\huge{ANSWER}$

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
• #Bal

NOTE:

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

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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

Answer 2

Solution :

A stone is projected from a point on the grd in a direction to hit a bird on the top of a telegraph post of height h.

The stone attains a maximum height of 2hm from the grd .

At the inst a nt of projection, the bird is flying away horizontally with an uniform speed.  

We have to find the ratio of horizontal velocities of the bird & the stone( assuming that the stone hits the bird)  

So

"  stone is projected from a point on the grd in a direction to hit a bird on the top of a telegraph post of height h. "

Suppose that the stone is projected with a velocity v and the angle made by it wrt the horizontal is ω

Maximum height attained is 2h

$\dfrac{v^2 sin^2 \omega }{2g} = 2h \\\implies v = \dfrac{\sqrt{4gh}}{sin \: \omega}$  

This value obtained will be useful later.

Equation of trajectory of a projectile :

$y = x tan\: \omega (1 - x/r) \:$  

We need to find t

Applying the second equation of motion along the y axis

y = ut + 1/2 at^2

h = vsinωt - 1/2gt^ 2

Substitute the initially obtained value of v here

$h = \sqrt{4gh} - 1/2 gt^2$

t has to be the variable

$1/2 gt^2 = \sqrt{4gh} - h \\t^2 = ( \sqrt{4gh} - h ) \multiply \dfrac{2}{g} \\t = \sqrt( ( \sqrt{4gh} - h ) \multiply \dfrac{2}{g} )$

Solving, we get

$t = \sqrt{\dfrac{h}{g} } (2\pm \sqrt{2})$

Ratio of horizontal velocities : 1 - (t/t')

= 1 - ( 2-root2)/(2+root2)

= 2/(2+root2)

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