A-gun-kept-on-a-straight-horizontal-road-is-used-to-hit-a-car-travelling-along-the-same-road-away-from-it-with-a-uniform-speed-of-72-km-h-The-car-is-at-a-distance-of-500-m-from-the-gun-when-the-gu

Question Number 224688 by Tawa11 last updated on 26/Sep/25

A gun, kept on a straight horizontal road, is used
to hit a car travelling along the same road away
from it with a uniform speed of 72 km/h. The car
is at a distance of 500 m from the gun, when the
gun is fired at an angle of 45° with the horizontal.
Find the distance of the car from the gun, when
the shell hits it.
g = 10m/s²

Answered by fantastic last updated on 26/Sep/25

Commented by fantastic last updated on 26/Sep/25

S=((u^2 sin (2×45^0 ))/(10))=(u^2 /(10))m (u^2 /(10))=500+x ...i time taken by car to go x m=(x/(20))s(v=20ms^(−1) ) so ucos (45^0 )×(x/(20))=500+x ((ux)/(20(√2)))=500+x...ii solving we will get x≈243.960 780 m Distance =500+x≈743.960 780 m

$${S}=\frac{{u}^{\mathrm{2}} \mathrm{sin}\:\left(\mathrm{2}×\mathrm{45}^{\mathrm{0}} \right)}{\mathrm{10}}=\frac{{u}^{\mathrm{2}} }{\mathrm{10}}{m} \\ $$$$\frac{{u}^{\mathrm{2}} }{\mathrm{10}}=\mathrm{500}+{x}\:…{i} \\ $$$${time}\:{taken}\:{by}\:{car}\:{to}\:{go}\:{x}\:{m}=\frac{{x}}{\mathrm{20}}{s}\left({v}=\mathrm{20}{ms}^{−\mathrm{1}} \right) \\ $$$${so} \\ $$$${u}\mathrm{cos}\:\left(\mathrm{45}^{\mathrm{0}} \right)×\frac{{x}}{\mathrm{20}}=\mathrm{500}+{x} \\ $$$$\frac{{ux}}{\mathrm{20}\sqrt{\mathrm{2}}}=\mathrm{500}+{x}…{ii} \\ $$$${solving}\:{we}\:{will}\:{get}\:{x}\approx\mathrm{243}.\mathrm{960}\:\mathrm{780}\:{m} \\ $$$${Distance}\:=\mathrm{500}+{x}\approx\mathrm{743}.\mathrm{960}\:\mathrm{780}\:{m} \\ $$

Commented by fantastic last updated on 28/Sep/25

$${do}\:{u}\:{understand}\:{everything}? \\ $$$${i}\:{answered}\:{very}\:{briefly} \\ $$

Commented by Tawa11 last updated on 28/Sep/25

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Answered by mr W last updated on 30/Sep/25

u_x =u_y =v V=72 km/h=20 m/s vt−((gt^2 )/2)=0 ⇒t=((2v)/g) 500+Vt=vt 500+20×((2v)/g)=((2v^2 )/g) 250+20×(v/g)=(v^2 /g) v^2 −20v−2500=0 ⇒v=10+(√(10^2 +2500))=10(1+(√(26))) d=vt=10(1+(√(26)))×((2×10(1+(√(26))))/(10)) =20(1+(√(26)))^2 ≈743.96 m u=(√2)v≈86 m/s ⇒too slow for gun fire

$${u}_{{x}} ={u}_{{y}} ={v} \\ $$$${V}=\mathrm{72}\:{km}/{h}=\mathrm{20}\:{m}/{s} \\ $$$${vt}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0}\:\Rightarrow{t}=\frac{\mathrm{2}{v}}{{g}} \\ $$$$\mathrm{500}+{Vt}={vt} \\ $$$$\mathrm{500}+\mathrm{20}×\frac{\mathrm{2}{v}}{{g}}=\frac{\mathrm{2}{v}^{\mathrm{2}} }{{g}} \\ $$$$\mathrm{250}+\mathrm{20}×\frac{{v}}{{g}}=\frac{{v}^{\mathrm{2}} }{{g}} \\ $$$${v}^{\mathrm{2}} −\mathrm{20}{v}−\mathrm{2500}=\mathrm{0} \\ $$$$\Rightarrow{v}=\mathrm{10}+\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{2500}}=\mathrm{10}\left(\mathrm{1}+\sqrt{\mathrm{26}}\right) \\ $$$${d}={vt}=\mathrm{10}\left(\mathrm{1}+\sqrt{\mathrm{26}}\right)×\frac{\mathrm{2}×\mathrm{10}\left(\mathrm{1}+\sqrt{\mathrm{26}}\right)}{\mathrm{10}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{20}\left(\mathrm{1}+\sqrt{\mathrm{26}}\right)^{\mathrm{2}} \approx\mathrm{743}.\mathrm{96}\:{m} \\ $$$$ \\ $$$${u}=\sqrt{\mathrm{2}}{v}\approx\mathrm{86}\:{m}/{s}\:\Rightarrow{too}\:{slow}\:{for}\:{gun}\:{fire} \\ $$

Commented by fantastic last updated on 30/Sep/25

$${there}\:{is}\:{nothing}\:{we}\:{can}\:{do}\:{type} \\ $$$${question} \\ $$

Commented by Tawa11 last updated on 01/Oct/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$

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