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NEET
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Physics
Chapter
Motion in Two Dimensions

NEET · Physics · Motion in Two Dimensions

A water fountain shoots a stream of water at an angle of 60^ to the...

A water fountain shoots a stream of water at an angle of 60∘60^{\circ} to the horizontal. If the maximum height reached by the water droplets is 15 m15 \text{ m}, what is the initial speed of the water? (Take g=10 m s−2g = 10 \text{ m s}^{-2})

Options

  1. A400 m s−1400 \text{ m s}^{-1}
  2. B203 m s−120\sqrt{3} \text{ m s}^{-1}
  3. C20 m s−120 \text{ m s}^{-1}
  4. D102 m s−110\sqrt{2} \text{ m s}^{-1}

Correct answer

C. 20 m s−120 \text{ m s}^{-1}

Step-by-step solution

The maximum height HH of a projectile is given by the formula:

H=u2sin⁡2θ2gH = \dfrac{u^2 \sin^2\theta}{2g}

Given that H=15 mH = 15 \text{ m}, θ=60∘\theta = 60^{\circ}, and g=10 m s−2g = 10 \text{ m s}^{-2}, we can substitute these values into the formula:

15=u2sin⁡2(60∘)2×1015 = \dfrac{u^2 \sin^2(60^{\circ})}{2 \times 10}

Since sin⁡(60∘)=32\sin(60^{\circ}) = \dfrac{\sqrt{3}}{2}, we have sin⁡2(60∘)=34\sin^2(60^{\circ}) = \dfrac{3}{4}.

15=u2×342015 = \dfrac{u^2 \times \dfrac{3}{4}}{20}

15×20=u2×3415 \times 20 = u^2 \times \dfrac{3}{4}

300=u2×34300 = u^2 \times \dfrac{3}{4}

u2=300×43=400u^2 = \dfrac{300 \times 4}{3} = 400

u=400=20 m s−1u = \sqrt{400} = 20 \text{ m s}^{-1}

Answer: 20 m s−120 \text{ m s}^{-1}
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