Question

Solve this with proper explanation.​

Answer 1

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$\boxed{ \fbox{horizontal \: range = \frac{ {u}^{2} {sin}^{2} \theta }{g} }} \\ \\ \boxed{ \fbox{maximum \: range = \frac{ {u}^{2} {sin}^{2} \theta }{2g} }}$

According to question:-

horizontal range=3×maximum height

$\frac{ {u}^{2} {sin}^{2} \theta }{g} = \frac{3 \times {u}^{2} { {sin}^{2} } \theta }{2g} \\ \\ \boxed{ \fbox{ {sin}^{2} \theta = 2sin.cos}} \ \\ \ \\ 2sin \theta.cos = \frac{3}{2} {sin}^{2} \theta \\ \\ tan \theta = \frac{4}{3} \\ \theta = {tan}^{ - 1} ( \frac{4}{3} )$

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second method

Let the angle of projection be @

Let maximum range be R and maximum height be H

According to given condition -> R=3H

Since formula for R is R = u2Sin2@/g and formula for H is H = u2Sin2@/2g

Therefore according to condition ----> u2Sin2@/g = 3 x u2Sin2@/2g

u2 , g will be cancelled

Since Sin2@ = 2Sin@Cos@

2Sin@Cos@ = 3 x Sin2@/2

2Cos@ = 3 x Sin@/2 (Sin@ and Sin2@ are cancelled)

4Cos@ = 3Sin@

tan@ = 4/3 ----> @ = tan-1(4/3)

Angle of projection is @ = tan-1(4/3)

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hops this may help you

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Answer 2

Answer:

According to question:-

horizontal range=3×maximum height

\begin{lgathered}\frac{ {u}^{2} {sin}^{2} \theta }{g} = \frac{3 \times {u}^{2} { {sin}^{2} } \theta }{2g} \\ \\ \boxed{ \fbox{ {sin}^{2} \theta = 2sin.cos}} \ \\ \ \\ 2sin \theta.cos = \frac{3}{2} {sin}^{2} \theta \\ \\ tan \theta = \frac{4}{3} \\ \theta = {tan}^{ - 1} ( \frac{4}{3} )\end{lgathered}

g

u

2

sin

2

θ

=

2g

3×u

2

sin

2

θ

sin

2

θ=2sin.cos

2sinθ.cos=

2

3

sin

2

θ

tanθ=

3

4

θ=tan

−1

(

3

4

)

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Hope it helps uhh

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