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Numerical · Q26

Q.A projectile is thrown at an angle of 30° to the horizontal. What should be the range of the initial velocity (u) so that its range will be between 40 m and 50 m? Assume g = 10 m/s².

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Given θ=30°, g=10 m/s², so sin⁡2θ=sin⁡60°=32≈0.8660\sin2\theta=\sin60°=\dfrac{\sqrt3}{2}\approx0.8660. From R=u2sin⁡2θgR=\dfrac{u^2\sin2\theta}{g}, u=Rgsin⁡2θu=\sqrt{\dfrac{Rg}{\sin2\theta}}. For R=40R=40 m: u=40×100.8660=461.98≈21.49u=\sqrt{\dfrac{40\times10}{0.8660}}=\sqrt{461.98}\approx21.49 m/s. For R=50R=50 m: u=50×100.8660=577.48≈24.03u=\sqrt{\dfrac{50\times10}{0.8660}}=\sqrt{577.48}\approx24.03 m/s. So the ini …

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