First slide

Projection Under uniform Acceleration

Question

An object is projected at angle (other than 90⁰) to the horizontal, from the ground. When it is at highest point of its path, magnitude of its position vector with respect to the point of projection is $\sqrt{2}$ times the maximum height reached by it. The angle of projection is

Moderate

A

$45^{0}$
B

$\tan^{- 1} (2)$
C

$\tan^{- 1} (4)$
D

$60^{0}$

Solution

$p o s i t i o n v e c t o r o f t h e h i g h e s t p o i n t \hat{i} \frac{R}{2} + \hat{j} H a n d i t s m a g n i t u d e i s \sqrt{\frac{R^{2}}{4} + H^{2}} g i v e n t h a t \sqrt{\frac{R^{2}}{4} + H^{2}} = \sqrt{2} H \frac{R^{2}}{4} + H^{2} = 2 H^{2} \frac{R}{2} = H \frac{H}{R} = \frac{1}{2} a n d \frac{H}{R} = \frac{1}{4} \tan θ = \frac{1}{2} \tan θ = 2$

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