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anonymous

  • one year ago

Would any one derive and explain the formula... Range of a projectle=2 height of a projectile. Any help would be greatly appreciated...

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  1. anonymous
    • one year ago
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    @irishboy123

  2. anonymous
    • one year ago
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    @arindameducationusc

  3. anonymous
    • one year ago
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    @michele_laino

  4. IrishBoy123
    • one year ago
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    can you scan or link the question pls :p

  5. IrishBoy123
    • one year ago
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    range R for generalised velocity v at angle \(\theta \) is given by: \(R = \dfrac{v^2 \sin 2 \theta} {g} \) the follows from equations of motion from \(v = u + at\), we have half flight time \(t_{1/2} \) is \(t_{1/2} = \dfrac{0 - v \sin \theta}{-g} = \dfrac{ v \sin \theta}{g}\) thus horizontal range R is \(R = 2 \times t_{1/2} \times v \cos \theta = 2 \dfrac{ v \sin \theta}{g} v \cos \theta = \dfrac{v^2 \sin 2 \theta} {g}\) the max height \(H_{max}\) for that generalised trajectory comes from \(v^2 = u^2 + 2ax\) and gives us \[H_{max} = \dfrac{v^2 \sin^2 \theta}{2g}\] if you are looking for a trajectory where \(R = 2 H_{max}\) then \(\dfrac{v^2 \sin 2 \theta} {g} = 2 \dfrac{v^2 \sin^2 \theta}{2g}\) \(\implies \sin 2 \theta = \sin^2 \theta \) \(\implies \tan \theta = 2 \)

  6. arindameducationusc
    • one year ago
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    I agree with @IrishBoy123 The question must be like this During a projectile motion if the max height equals horizontal rangle, then the angle of projection with the horizontal is? And @IrishBoy123 has given the answer

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