A bullet fired from a gun which accute maximum highet 25m and its horizontal range 200.calculate the angle of prozention of bullet.
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It should be “……which acquires maximum height……”
Maximum height = 25 m
u² sin²θ / (2g) = 25 m
(u sinθ)² = 2 g × 25
(u sinθ) = √(2g × 25)
= √(2 × 10 × 25)
= 10√5 m/s …………[1]
Range = 200 m
2u²sinθ cosθ / g = 200
Rearrange the above equation
ucosθ = 200 g / (2u sinθ)
= 200 × 10 / (2 × 10√5)
= 100/√5 m/s …………[2]
Divide equation (1) by (2)
usinθ / (ucosθ) = [10√5] / [100/√5]
Tanθ = 1/2
θ = Tan⁻¹ (1/2)
Angle of projection of bullet is Tan⁻¹ (1/2)
Maximum height = 25 m
u² sin²θ / (2g) = 25 m
(u sinθ)² = 2 g × 25
(u sinθ) = √(2g × 25)
= √(2 × 10 × 25)
= 10√5 m/s …………[1]
Range = 200 m
2u²sinθ cosθ / g = 200
Rearrange the above equation
ucosθ = 200 g / (2u sinθ)
= 200 × 10 / (2 × 10√5)
= 100/√5 m/s …………[2]
Divide equation (1) by (2)
usinθ / (ucosθ) = [10√5] / [100/√5]
Tanθ = 1/2
θ = Tan⁻¹ (1/2)
Angle of projection of bullet is Tan⁻¹ (1/2)
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