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Standard XII

Physics

Acceleration in 2D

A ball is pro...

Question

A ball is projected with a velocity $20 m s^{- 1}$ at an angle to the horizontal, In order to have the maximum range. Its velocity at the highest position must be

10 m s^{- 1}

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14 m s^{- 1}

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18 m s^{- 1}

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16 m s^{- 1}

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Solution

The correct option is B $14 m s^{- 1}$
Range= $\frac{u^{2} sin 2 θ}{2 g}$
Thus maximum range occurs when $sin 2 θ = 1 o r θ = 45^{0}$
Thus velocity at highest position is only horizontal $= u cos θ = u cos 45^{0} = 20 / \sqrt{2}$ , which is nearly 14m/s

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A ball is projected with a velocity 20 ms−1 at an angle to the horizontal, In order to have the maximum range. Its velocity at the highest position must be

The correct option is B 14 ms−1Range=u2sin2θ2gThus maximum range occurs when sin2θ=1 orθ=450Thus velocity at highest position is only horizontal =ucosθ=ucos450=20/√2, which is nearly 14m/s

A ball is projected with a velocity $20 m s^{- 1}$ at an angle to the horizontal, In order to have the maximum range. Its velocity at the highest position must be

The correct option is B $14 m s^{- 1}$
Range= $\frac{u^{2} sin 2 θ}{2 g}$
Thus maximum range occurs when $sin 2 θ = 1 o r θ = 45^{0}$
Thus velocity at highest position is only horizontal $= u cos θ = u cos 45^{0} = 20 / \sqrt{2}$ , which is nearly 14m/s