Question

The horizontal range is four times the maximum height attained by a projectile. The angle of projection is [MP PET 1994; CBSE PMT 2000; RPET 2001]
A) {{90}^{o}} B) {{60}^{o}} C) {{45}^{o}} D) {{30}^{o}}

Answer 1

yeah sure for your help


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Here is your answer

option C is definitely ur correct answer

45 degree is the angle of projection

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Answer 2

Answer:

Angle of projection is 45°

Explanation:

It is given that, the horizontal range is four times the maximum height attained by a projectile.

The mathematical expression for the horizontal range is given by :

$R=\dfrac{v^2\ sin\ 2\theta}{g}$

Where

v is the velocity of projectile

$\theta$ is the angle of projection

The mathematical expression for the maximum height is given by :

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

According to given condition:

R = 4 (maximum height)

$\dfrac{v^2\ sin\ 2\theta}{g}=4\times \dfrac{v^2\ sin^2\ \theta}{2g}$

$sin\ 2\theta=2sin^2\ \theta$

$\theta=45^\circ$

On solving the angle of projection becomes 45°. Hence, this is the required solution.