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The ceiling of a hall is 40m  high. For a maximum horizontal distance, the angle at which the ball may be thrown

with a speed of 56m{{s}^{-1}}  without hitting the ceiling of the hall is

Option: 1

{{25}^{\circ }}


Option: 2

{{30}^{\circ }}


Option: 3

{{45}^{\circ }}


Option: 4

{{60}^{\circ }}


Answers (1)

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Answer : B) {{30}^{\circ }}

Explanation : Here, u=56m{{s}^{-1}}

Let \theta be the angle of projection with the horizontal to have maximum range, with maximum height, H=40m

Maximum Height, H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}

H=\frac{{{(56)}^{2}}{{\sin }^{2}}\theta }{2\times 9.8}

40=\frac{{{(56)}^{2}}{{\sin }^{2}}\theta }{2\times 9.8}

{{\sin }^{2}}\theta =\frac{2\times 9.8\times 40}{{{(56)}^{2}}}

{{\sin }^{2}}\theta =\frac{1}{4}

\sin \theta =\frac{1}{2}

\theta ={{\sin }^{-1}}(\frac{1}{2})

\theta ={{30}^{\circ }}

 

Posted by

Pankaj Sanodiya

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