SATHEE: Motion in a Plane - Result Question 31

Motion in a Plane - Result Question 31

34. The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectiles is :

[2012]

(a) $θ = \tan^{- 1} (\frac{1}{4})$

(b) $θ = \tan^{- 1}$

(d) $θ = 45^{\circ}$

Show Answer

Solution:

(b) Horizontal range,

$R = \frac{u^{2} \sin 2 θ}{g}$

Maximum height,

$H = \frac{u^{2} \sin^{2} θ}{2 g}$

According to the condition,

$R = H$

$\Rightarrow \frac{u^{2} \sin 2 θ}{g} = \frac{u^{2} \sin^{2} θ}{2 g}$

$\Rightarrow 2 \sin θ \cos θ = \frac{\sin^{2} θ}{2}$

$2 \cos θ = \frac{\sin θ}{2}$

$\Rightarrow \cot θ = \frac{1}{4}$

$\Rightarrow \tan θ = 4$

$\Rightarrow θ = [\tan^{- 1} (4)]$

In case of projective motion range $R$ is $n$ times the maximum height $H$

i.e., $R = n H \Rightarrow \frac{u^{2} \sin 2 θ}{g} = n \frac{u^{2} \sin^{2} θ}{2 g}$

$\Rightarrow T a n θ = \frac{4}{n}$ Angle of projection $θ = \tan^{- 1} (\frac{4}{n})$

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Motion in a Plane - Result Question 31

34. The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectiles is :

Solution:

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