Question

So, Initial speed of bullet be 'v'. ( t=frac{2 v sin (theta)}{g} )
2) We also observe that,
Speed of car, ( u=72 sqrt{2} mathrm{km} / mathrm{h}=2 mathrm{O} sqrt{2} mathrm{m} / mathrm{s} )
Distance covered by car Range of Projectile ( =50 mathrm{m}+20 sqrt{2}^{*} mathrm{t} )
( =>frac{v^{2} sin (2 times 45)}{g}=50+20 sqrt{2}left(frac{2 v sin (45)}{g}right) )
( =>v^{2}-40 v-500=0 )
( =>(v-50)(v+10)=0 )
( =>v=50 quad(v eq-10) )
(b) Hence, Speed of Projection of shell ( =50 mathrm{m} / mathrm{s} )
(a) Distance from the gun:
Range of Projectile:
( =frac{v^{2} sin (2 times 45)}{g}=frac{50^{2} times 1}{10}=250 mathrm{m} )
Hence, Distance from gun ( =250 mathrm{m} )

# A-8. WISE UNUN) A gun kept on a straight horizontal road is used to hit a car, travelling along the same road away from the gun with a uniform speed of 72 x 12 km/hour. The car is at a distance of 50 metre from the gun, when the gun is fired at an angle of 45° with the horizontal. Find (i) the distance of the car from the gun when the shell hits it, (ii) the speed of projection of the shell from the gun. (g = 10 m/s?] [IIT 1974)

Solution