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A man standing on a hill top projects a stone horizontally with speed v0 as shown in figure. Taking the coordinate system as given in the figure. The coordinates of the point where the stone will hit the hill surface |
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Answer» Answer: (2V₀²Tanθ/g , - 2V₀²Tan²θ/g) A man STANDING on a hill top projects a stone horizontally with speed v0 as shown in FIGURE. Taking the coordinate system as GIVEN in the figure. The coordinates of the point where the stone will hit the hill surface Horizontal speed = V₀ Vertical Speed = 0 Let say coordinate where stone hit the surface = ( x , -y) S = ut + (1/2)at² y = (1/2)gt² t² = 2y/g t = √(2y/g) Horizontal distance = x = V₀ * √(2y/g) TAN θ = y / (V₀ * √(2y/g)) => Tanθ = √y √g / V₀ * √2 => √y = V₀ * √2Tanθ/√g Squaring both sides => y = 2V₀²Tan²θ/g x = V₀ * √(2y/g) = V₀ * √ ( 2 * 2V₀²Tan²θ/g²) => x = V₀ * 2V₀Tanθ/g => x = 2V₀²Tanθ/g (2V₀²Tanθ/g , - 2V₀²Tan²θ/g) |
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