A bullet fired from rifle attains a maximum height of 5m and crosses a range of 200m.find the angle of projection
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maximum height = u²sin²θ/2g=5 ---(1)
range=u²sin2θ/g=200 ---(2)
dividing (1) by (2)
(u²sin²θ/2g) / (u²sin2θ/g) = 5/200 {u²will cancel u² and g wiil cancel g}
(sin²θ/2) / (sin2θ)=1/40
(sinθ×sinθ/2) / (2sinθcosθ) = 1/40 {sinθ will cancel sinθ}
(sinθ/2) / (2cosθ)=1/40
40sinθ=4cosθ
tanθ=1/10
θ=tan-¹ 1/10 (tan-¹ means tan inverse)
range=u²sin2θ/g=200 ---(2)
dividing (1) by (2)
(u²sin²θ/2g) / (u²sin2θ/g) = 5/200 {u²will cancel u² and g wiil cancel g}
(sin²θ/2) / (sin2θ)=1/40
(sinθ×sinθ/2) / (2sinθcosθ) = 1/40 {sinθ will cancel sinθ}
(sinθ/2) / (2cosθ)=1/40
40sinθ=4cosθ
tanθ=1/10
θ=tan-¹ 1/10 (tan-¹ means tan inverse)
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