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IV. Exercises · Q6

Q.An object is projected at an angle such that the horizontal range is 4 times the maximum height. What is the angle of projection of the object?

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Step 1. R=u2sin⁡2θgR=\dfrac{u^2\sin2\theta}{g}, H=u2sin⁡2θ2gH=\dfrac{u^2\sin^2\theta}{2g}. Given R=4HR=4H.

Step 2. u2sin⁡2θg=4×u2sin⁡2θ2g=2u2sin⁡2θg\dfrac{u^2\sin2\theta}{g}=4\times\dfrac{u^2\sin^2\theta}{2g}=\dfrac{2u^2\sin^2\theta}{g}. …

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