A projectile is projected with initial velocity -6 -8 -m-sec-If-g-10-ms-2- then horizontal range is 4-8 metre-9-6 metre-19-2 metre-14-0 metre-

The correct option is B 9.6 metre Initial velocity (6 î+8 Ĵ)m/s given Magnitude of velocity of projection u=√(u^2_x+u^2_y)=√(6^2+8^2)=10 m/s Angle of projectio ...

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Standard XII

Physics

Range

A projectile ...

Question

A projectile is projected with initial velocity $(6^i + 8^j) m / s e c . I f g = 10 m s^{- 2}$ , then horizontal range is

4.8 metre

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9.6 metre

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19.2 metre

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14.0 metre

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Solution

The correct option is B
9.6 metre

Initial velocity $(6^i + 8^J) m / s$ given

Magnitude of velocity of projection $u = \sqrt{u_{x}^{2} + u_{y}^{2}} = \sqrt{6^{2} + 8^{2}} = 10 m / s$

Angle of projection $t a n θ = \frac{u_{y}}{u_{x}} = \frac{8}{6} = \frac{4}{3} ∴ s i n θ = \frac{4}{5} a n d c o s θ = \frac{3}{5}$

Now horizontal range $R = \frac{u^{2} s i n 2 θ}{g} = \frac{u^{2} 2 s i n θ c o s θ}{g} = \frac{(10)^{2} \times 2 \times \frac{4}{5} \times \frac{3}{5}}{10} = 9.6 m e t e r$

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A projectile is projected with initial velocity (6^i+8^j)m/sec. If g=10 ms−2, then horizontal range is

The correct option is B 9.6 metre Initial velocity (6^i+8^J)m/s given Magnitude of velocity of projection u=√u2x+u2y=√62+82=10 m/s Angle of projection tan θ=uyux=86=43 ∴ sin θ=45 and cos θ=35 Now horizontal range R=u2 sin 2 θg=u2 2 sinθ cos θg=(10)2×2×45×3510=9.6 meter

A projectile is projected with initial velocity $(6^i + 8^j) m / s e c . I f g = 10 m s^{- 2}$ , then horizontal range is