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Standard XII

Physics

Horizontal and Vertical Components of Projectile Motion

The maximum h...

Question

The maximum horizontal distance covered by the ball will be? If angle of projection is $30^{\circ}$ .

160 \sqrt{3} m

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140 \sqrt{3} m

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120 \sqrt{3} m

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100 \sqrt{3} m

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Solution

The correct option is C $160 \sqrt{3} m$
$R = \frac{u^{2} sin 2 θ}{g} = \frac{(56)^{2} sin 60^{\circ}}{9.8}$
$= \frac{56 \times 56 \times \sqrt{3}}{19.6} = 160 \sqrt{3} m$ .

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The maximum horizontal distance covered by the ball will be? If angle of projection is 30∘.

The correct option is C 160√3mR=u2sin2θg=(56)2sin60∘9.8=56×56×√319.6=160√3m.

The maximum horizontal distance covered by the ball will be? If angle of projection is $30^{\circ}$ .

The correct option is C $160 \sqrt{3} m$
$R = \frac{u^{2} sin 2 θ}{g} = \frac{(56)^{2} sin 60^{\circ}}{9.8}$
$= \frac{56 \times 56 \times \sqrt{3}}{19.6} = 160 \sqrt{3} m$ .