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Question

Single Correct Answer Type
A projectile initially has the same horizontal velocity as it would acquire, if it had moved from rest with uniform acceleration of $$3  ms^{-2}$$ for $$0.5  min$$. If the maximum height reached by it is $$80  m$$, then the angle of projection is $$(g = 10  ms^{-2})$$


A
tan1(3)
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B
tan1(32)
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C
tan1(49)
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D
sin1(49)
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Solution

The correct option is C $$ tan ^{-1}(\dfrac{4}{9})$$
$$H = \dfrac {u^{2}\sin^{2}\theta}{2g}$$ 

or $$80 = \dfrac {u^{2}\sin^{2}\theta}{2\times 10}$$

or $$u^{2} \sin^{2}\theta = 1600$$ 

or $$u\sin \theta = 40\ ms^{-1}$$

$$\dfrac {u\sin \theta}{u\cos \theta} = \dfrac {40}{90}$$

 or $$\tan \theta = \dfrac {4}{9}$$

 or $$\theta = \tan^{-1} \left (\dfrac {4}{9}\right )$$.

Physics

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