A projectile is projected with initial velocity 6- - 8 - m-s- If g - 10 ms-2- then horizontal range is-

The correct option is A 9.6 metre vsinθ = 8 vcosθ =6 horizontal range is R = v^2sin2θg = 2vcos θ v sinθ10 = 2× 6 × 810 = 9.6 #160;metre ...

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Work Energy Theorem Application

A projectile ...

Question

A projectile is projected with initial velocity $(6^i + 8^j)$ m/s. If $g = 10$ $m s^{- 2}$ , 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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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

x = 4 t

y = 3 t - 4 t^{2}

\sqrt{2}

u = 30 m/s

90 m

(

g = 10 {m/s}^{2}

)

(6^i + 8^j) m s^{- 1}

g = 10 m s^{- 2}

\sqrt{2}

u = 30 m/s

90 m

(

g = 10 {m/s}^{2}

)

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