SATHEE: Motion in a Plane - Result Question 33

Motion in a Plane - Result Question 33

36. A projectile is fired at an angle of $45^{\circ}$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is

[2011M]

(a) $60^{\circ}$

(b) $\tan^{- 1} (\frac{1}{2})$

(d) $45^{\circ}$

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Answer:

Correct Answer: 36. (b)

Solution:

(b) Maximum height,

$H = \frac{u^{2} \sin^{2} 45^{\circ}}{2 g} = \frac{u^{2}}{4 g}$

Range, $R = \frac{u^{2} \sin 90^{\circ}}{g} = \frac{u^{2}}{g}$

$∴ \frac{R}{2} = \frac{u^{2}}{2 g}$

According to the required condition,

$∴ \tan α = \frac{H}{R / 2}$

Put the volues from equation (1) and (2)

$\Rightarrow \frac{(\frac{u^{2}}{4 g})}{(\frac{u^{2}}{2 g})} \tan α ∴ α = \tan^{- 1} (\frac{1}{2})$

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Motion in a Plane - Result Question 33

36. A projectile is fired at an angle of 45∘ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is

Answer:

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36. A projectile is fired at an angle of $45^{\circ}$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is