Question

plz help ^_^


kindly don't spam :'(​

Answer 1

★ Concept :-

Here the concept of Maximum height and Horizontal Range of a projectile launched by some angle with horizontal has been used. We see that R shows the Horizontal Range of the projectile and hm shows the maximum height of the projectile. So firstly we can apply the required values in the formula and then find the ratio of Maximum Height and Horizontal Range and thus get the answer.

Let's do it !!

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★ Formula Used :-

$\;\boxed{\sf{\pink{h_{m}\;=\;\bf{\dfrac{u\sin^{2}\theta}{2g}}}}}$

$\;\boxed{\sf{\pink{R\;=\;\bf{\dfrac{u\sin^{2}2\theta}{g}}}}}$

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★ Solution :-

Given,

» Angle of projection = θ = θ｡ (according to the qúestion)

• Let the initial velocit the of the projectile be u

• Let the acceleration due to gravity be g

• Maximum height of the projectile = hm

• Horizontal range of projectile = R

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~ For the value of hm and R ::

We know that,

$\;\sf{\rightarrow\;\;h_{m}\;=\;\bf{\dfrac{u\sin^{2}\theta}{2g}}}$

By applying values in this, we get

$\;\bf{\rightarrow\;\;\red{h_{m}\;=\;\bf{\dfrac{u\sin^{2}\theta_{0}}{2g}}}}$

Also,

$\;\sf{\rightarrow\;\;R\;=\;\bf{\dfrac{u\sin^{2}2\theta}{g}}}$

By applying values in this, we get

$\;\sf{\rightarrow\;\;\green{R\;=\;\bf{\dfrac{u\sin^{2}2\theta_{0}}{g}}}}$

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~ For the final answer ::

Now taking the ratio of Maximum Height by Horizontal Range, we get

$\;\bf{\Longrightarrow\;\;\blue{\dfrac{h_{m}}{R}\;=\;\dfrac{\frac{u\sin^{2}\theta_{0}}{2g}}{\frac{u\sin^{2}2\theta_{0}}{g}}}}$

Cancelling g, we get

$\;\bf{\Longrightarrow\;\;\dfrac{h_{m}}{R}\;=\;\dfrac{\frac{u\sin^{2}\theta_{0}}{2}}{u\sin^{2}2\theta_{0}}}$

$\;\bf{\Longrightarrow\;\;\dfrac{h_{m}}{R}\;=\;\dfrac{u\sin^{2}\theta_{0}}{2u\sin^{2}2\theta_{0}}}$

Cancelling u we get,

$\;\bf{\Longrightarrow\;\;\dfrac{h_{m}}{R}\;=\;\dfrac{\sin^{2}\theta_{0}}{2\sin^{2}2\theta_{0}}}$

We know that, sin² 2θ = 2cos θ sin θ

By applying this, we get

$\;\bf{\Longrightarrow\;\;\dfrac{h_{m}}{R}\;=\;\dfrac{\sin^{2}\theta_{0}}{2(2\cos\theta_{0}\sin\theta_{0})}}$

Cancelling sin θ｡ we get,

$\;\bf{\Longrightarrow\;\;\dfrac{h_{m}}{R}\;=\;\dfrac{\sin\theta_{0}}{4\cos\theta_{0}}}$

$\;\bf{\Longrightarrow\;\;\dfrac{h_{m}}{R}\;=\;\dfrac{1}{4}\:\times\:\dfrac{\sin\theta_{0}}{\cos\theta_{0}}}$

$\;\bf{\Longrightarrow\;\;\dfrac{h_{m}}{R}\;=\;\dfrac{1}{4}\:\times\:\tan\theta_{0}}$

$\;\bf{\Longrightarrow\;\;\dfrac{1}{4}\:\times\:\tan\theta_{0}\;=\;\dfrac{h_{m}}{R}}$

$\;\bf{\Longrightarrow\;\;\tan\theta_{0}\;=\;\dfrac{4h_{m}}{R}}$

$\;\bf{\Longrightarrow\;\;\tan\theta_{0}\;=\;\bigg(\dfrac{4h_{m}}{R}\bigg)}$

$\;\bf{\Longrightarrow\;\;\purple{\theta_{0}\;=\;\tan^{-1}\bigg(\dfrac{4h_{m}}{R}\bigg)}}$

This is the required answer.

Hence proved.

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★ More to know :-

$\;\tt{\leadsto\;\;Time\;of\;Flight\;=\;\dfrac{2u\sin\theta}{g}}$

$\;\tt{\leadsto\;\;R_{max}\;=\;\dfrac{u^{2}}{g}}$

$\;\tt{\leadsto\;\;\theta\;=\;\dfrac{Arc}{Radius}}$

$\;\tt{\leadsto\;\; Centripetal\; Acceleration\;=\;\dfrac{v^{2}}{r}}$

$\;\tt{\leadsto\;\;v\;=\;r\omega}$