Question

A body is projected at such an angle that the horizontal range is three times the greatest height. Then the angle projection is ...

Answer 1

Hey friend, Harish here.

Here is your answer.

WE KNOW THAT :

$\mathrm{ 1) \ Range = \frac{u^2 \sin 2\theta}{g} } \\ \\ \mathrm{ 2) \ Max \ Height = \frac{u^2 \sin^2 \theta}{2g} }$

Given :

3 × Max. Height = Range

⇒ $\mathrm{3\times \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin 2 \theta}{g}}$

NOTE :   Sin 2θ = 2. Sinθ. Cosθ

⇒ $\mathrm{3\times \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 . 2.\sin \theta.\cos \theta}{g}}$

⇒ $\mathrm{2 \cos \theta = \frac{3}{2} \sin \theta }$

∴ $\boxed{\theta = \tan ^{-1} \Big(\frac{4}{3}\Big) = 53^{\cir} }$

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Hope my answer is helpful to you.

Answer 2

here is your answer OK ☺☺☺☺☺☺☺



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)