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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)

best_answer

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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