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Standard XII

Physics

Equations of Motion

A ball is thr...

Question

A ball is thrown with initial velocity of 10 m/s at an angle of 23 $^{o}$ with horizontal surface. How far away does the ball land? Given that sin 23 $^{o}$ = 0.4, cos 23 $^{o}$ = 0.9, and g = $10 m / s^{2}$ and ignore the effect of air resistance.

3.6 m

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4.0 m

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4.5 m

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7.2 m

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9.0 m

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Solution

The correct option is A 7.2 m
Given : $u = 10$ m/s $θ = 23^{o}$ $g = 10 m / s^{2}$
Horizontal range of particle $R = \frac{u^{2} s i n 2 θ}{g} = \frac{2 u^{2} s i n θ c o s θ}{g} = \frac{2 \times 10^{2} \times 0.4 \times 0.9}{10} = 7.2$ m

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A ball is thrown with initial velocity of 10 m/s at an angle of 23o with horizontal surface. How far away does the ball land? Given that sin 23o = 0.4, cos 23o = 0.9, and g = 10m/s2and ignore the effect of air resistance.

The correct option is A 7.2 mGiven : u=10 m/s θ=23o g=10m/s2Horizontal range of particle R=u2sin2θg=2u2sinθcosθg=2×102×0.4×0.910=7.2 m

A ball is thrown with initial velocity of 10 m/s at an angle of 23 $^{o}$ with horizontal surface. How far away does the ball land? Given that sin 23 $^{o}$ = 0.4, cos 23 $^{o}$ = 0.9, and g = $10 m / s^{2}$ and ignore the effect of air resistance.

The correct option is A 7.2 m
Given : $u = 10$ m/s $θ = 23^{o}$ $g = 10 m / s^{2}$
Horizontal range of particle $R = \frac{u^{2} s i n 2 θ}{g} = \frac{2 u^{2} s i n θ c o s θ}{g} = \frac{2 \times 10^{2} \times 0.4 \times 0.9}{10} = 7.2$ m