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The-range-of-riffle-bullet-is-1000m-when-is-the-angle-of-projection-if-the-bullet-is-fired-with-the-same-angle-from-a-car-travelling-at-36km-h-towards-the-target-show-that-the-range-will-be-incre

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Question Number 50915 by peter frank last updated on 22/Dec/18
The range of riffle bullet is 1000m when θ is the angle of projection.if the bullet is fired with the same angle from a car travelling at 36km/h towards the target show that the range will be increased by ((1000)/7).(√(tanθ)) m.
$${The}\:{range}\:{of}\:{riffle}\:{bullet} \\ $$$${is}\:\mathrm{1000}{m}\:{when}\:\theta\:{is}\:{the}\:{angle} \\ $$$${of}\:{projection}.{if}\:{the}\:{bullet} \\ $$$${is}\:{fired}\:\:{with}\:{the}\:{same}\: \\ $$$${angle}\:\:{from}\:{a}\:{car}\:{travelling} \\ $$$${at}\:\mathrm{36}{km}/{h}\:{towards}\:{the}\:{target} \\ $$$${show}\:{that}\:{the}\:{range}\:{will} \\ $$$${be}\:{increased}\:{by} \\ $$$$\frac{\mathrm{1000}}{\mathrm{7}}.\sqrt{{tan}\theta}\:{m}. \\ $$
Answered by mr W last updated on 22/Dec/18
case 1: fired from the ground u sin θ t−(1/2)gt^2 =0 ⇒ t=((2 u sin θ)/g) L_1 =u cos θ t=((2 u^2 sin θ cos θ)/g)=((u^2 sin 2θ )/g) ⇒u=(√((gL_1 )/(sin 2θ))) case 2: fired from the car with speed v u sin θ t−(1/2)gt^2 =0 ⇒ t=((2 u sin θ)/g) L_2 =(u cos θ+v) t=u cos θ t +vt=L_1 +vt=L_1 +((2 uv sin θ)/g) ΔL=L_2 −L_1 =((2 uv sin θ)/g) ΔL=((2v sin θ)/g)(√((gL_1 )/(sin 2θ))) ΔL=v(√((4 sin^2 θ gL_1 )/(g^2 2 sin θ cos θ))) ⇒ΔL=v(√((2 tan θ L_1 )/g)) with v=36km/h=10 m/s, g=9.81m/s^2 , L_1 =1000 m ⇒ΔL=10(√((2 tan θ 1000)/(9.81))) ⇒ΔL=((1000)/( (√(49.05))))(√(tan θ))≈((1000)/7)(√(tan θ))
$${case}\:\mathrm{1}:\:{fired}\:{from}\:{the}\:{ground} \\ $$$${u}\:\mathrm{sin}\:\theta\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{t}=\frac{\mathrm{2}\:{u}\:\mathrm{sin}\:\theta}{{g}} \\ $$$${L}_{\mathrm{1}} ={u}\:\mathrm{cos}\:\theta\:{t}=\frac{\mathrm{2}\:{u}^{\mathrm{2}} \:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{{g}}=\frac{{u}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\theta\:}{{g}} \\ $$$$\Rightarrow{u}=\sqrt{\frac{{gL}_{\mathrm{1}} }{\mathrm{sin}\:\mathrm{2}\theta}} \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:{fired}\:{from}\:{the}\:{car}\:{with}\:{speed}\:{v} \\ $$$${u}\:\mathrm{sin}\:\theta\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{t}=\frac{\mathrm{2}\:{u}\:\mathrm{sin}\:\theta}{{g}} \\ $$$${L}_{\mathrm{2}} =\left({u}\:\mathrm{cos}\:\theta+{v}\right)\:{t}={u}\:\mathrm{cos}\:\theta\:{t}\:+{vt}={L}_{\mathrm{1}} +{vt}={L}_{\mathrm{1}} +\frac{\mathrm{2}\:{uv}\:\mathrm{sin}\:\theta}{{g}} \\ $$$$\Delta{L}={L}_{\mathrm{2}} −{L}_{\mathrm{1}} =\frac{\mathrm{2}\:{uv}\:\mathrm{sin}\:\theta}{{g}} \\ $$$$\Delta{L}=\frac{\mathrm{2}{v}\:\mathrm{sin}\:\theta}{{g}}\sqrt{\frac{{gL}_{\mathrm{1}} }{\mathrm{sin}\:\mathrm{2}\theta}} \\ $$$$\Delta{L}={v}\sqrt{\frac{\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:{gL}_{\mathrm{1}} }{{g}^{\mathrm{2}} \:\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}} \\ $$$$\Rightarrow\Delta{L}={v}\sqrt{\frac{\mathrm{2}\:\mathrm{tan}\:\:\theta\:{L}_{\mathrm{1}} }{{g}}} \\ $$$${with}\:{v}=\mathrm{36}{km}/{h}=\mathrm{10}\:{m}/{s},\:{g}=\mathrm{9}.\mathrm{81}{m}/{s}^{\mathrm{2}} ,\:{L}_{\mathrm{1}} =\mathrm{1000}\:{m} \\ $$$$\Rightarrow\Delta{L}=\mathrm{10}\sqrt{\frac{\mathrm{2}\:\mathrm{tan}\:\:\theta\:\mathrm{1000}}{\mathrm{9}.\mathrm{81}}} \\ $$$$\Rightarrow\Delta{L}=\frac{\mathrm{1000}}{\:\sqrt{\mathrm{49}.\mathrm{05}}}\sqrt{\mathrm{tan}\:\theta}\approx\frac{\mathrm{1000}}{\mathrm{7}}\sqrt{\mathrm{tan}\:\theta} \\ $$
Commented by peter frank last updated on 22/Dec/18
thank you
$${thank}\:{you} \\ $$

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