Question

Two balls are projected from the same points in the direction 30° and 60° with respective the horizontal. What is the ration of there initial velocities if they.
a)Atain the same height .
b)Atain the same height have same range .​

Answer 1

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Question:-

Two balls are projected from the same point in the direction 30° and 60° with respective the horizontal.What is the ratio of there initial velocities if they

a)Atain the same height

b)Atain ghe same height have same range.

Answer:-

Given,

• Two balls are projected from the same point in the directions 30° and 60°

So,

$\implies$ $\theta_{1}$ = 30°

$\implies$ $\theta_{2}$ = 60°

To Find,

• The ratio of there initial velocities in the given conditions

Calculations,

Here in the given question ,

According to first condition we need to find the ration of height.

A) Attain the same height :-

According to the maximum height formula

We know that,

H= $\dfrac { {u}^{2}{sin}^{2}\theta} {2g}$

H1 = $\dfrac{ {U_{1}}^{2} {sin}^{2} \theta}{2g}$

H2 = $\dfrac{ {U_{2} }^{2}{sin}^{2}\theta}{2g}$

Now,

H1 = H2

$\implies$ $\dfrac{ {U_{1} }^{2} {sin}^{2} \theta}{2g}$ = $\dfrac{ {U_{2} }^{2}{sin}^{2}\theta}{2g}$

Therefore , 2g gets canceled on both sides

$\implies$ $\dfrac{ {u_{1}}^{2} }{ {U_{2} }^{2} }$ = $\dfrac { {sin}^{2}\theta_{2} }{ {sin}^{2}\theta_{1}}$

Now,

Susbustitue the given theta values:-

$\implies$ $\dfrac{ {u_{1}}^{2} }{ {U_{2}}^{2} }$ = $\dfrac { {sin}^{2} 60° }{ {sin}^{2} 30°}$

$\implies$ $\dfrac{ u_{1} ^{2} }{ u_{2} ^{2} } = ( \dfrac{ \dfrac{ \sqrt{3} }{2} }{ \dfrac{1}{2} } ) ^{2}$

$\implies$ $\dfrac{ u_{1} ^{2} }{ u_{2} ^{2} } = \dfrac{ \dfrac{3}{4} }{ \dfrac{1}{4} }$

Here , 4 gets cancel on both sides.

so,

$\implies$ $= \dfrac{3}{1}$

$u_{ 1 } . u_{2} \: = \sqrt{3} .1$

B) Having the same range:-

According to maximum height,

Range R = $\dfrac{ {u}^{2}{sin2\theta} }{g}$

$R_{1}$ = $\dfrac{ {u_{1} }^{2}{sin2\theta}}{g}$

$R_{2}$ = $\dfrac{ {u_{2}}^{2}{sin2\theta}} {g}$

R1 = R2

$\implies$ $\dfrac { {u_{1} }^{2}{sin2\theta}} {g}$ = $\dfrac { {u_{2}}^{2}{sin2\theta}} {g}$

Here , g gets cancel on both side

$\implies$ $\dfrac{ u_{1} ^{2} }{ u_{2}^{2} }$= $\dfrac{Sin2\theta_{2}}{sin2\theta_{1}}$

$\implies$ $\dfrac{ u_{1} ^{2} }{ u_{2}^{2} }$ = $\dfrac{Sin2 (60)}{sin2 (30)}$

$\implies$ $\dfrac{ u_{1} ^{2} }{ u_{2}^{2} } = \frac{ \dfrac{ \sqrt{3} }{2} }{ \frac{ \sqrt{3} }{2} }$

Here root 3 and 2 gets cancel on both sides,

$\implies$ $\dfrac{ u_{1} ^{2} }{ u_{2} ^{2} } = \dfrac{1}{1}$

$u_{1}. u_{2} = 1.1$

So,

The ratio of the first condition (Attain the same height ) is $u_{ 1 } . u_{2} \: = \sqrt{3} .1$

The ratio of the second condition (Having the same range) is $u_{1}. u_{2} = 1.1$

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-Happies!!