A gun is kept on a horizontal road and is used to hit a running car. The uniform speed of the car is 72km/h . At the instant of firing at an angle of `45^@` with the horizontal , the car is at a distance of 500 m from the gun. Find out the distance between the gun and the car at the instant of hitting. Given, `g=10m//s^2`.

Explanation

Velocity of the car, `v=72 km //h=20 m//s`,
if it hit after a time t s , then its displacement =20t m
`therefore` Distance between the gun and the car at that instant, D=500 +20 t m.
If u be the initial velocity of the bullet, then its horizontal range,
`R=(u^2 sin (2xx45^@))/(g)=(u^2)/(g)`
`therefore (u^2)/(g) =500+20 t " " ....(1)`
Considering the vertical motion of the bullet in time t, we have
`0= u sin 45^@ *t-1/2"gt"^2 or, (ut)/(sqrt(2)) =1/2 "gt"^2`
or, `u=1/sqrt(2) "gt"`
`therefore u^2=1/2g^2t^2 or, (u^2)/(g)=1/2 "gt"^2 =1/2 xx10 xxt^2 =5t^2`
Putting in (1) ,we get
`5t^2=500+20t or, t^2-4t -100=0`
`therefore t=(4pmsqrt(16-4xx1xx(-100)))/(2xx1)=2pm sqrt(104)`
Keeping only the positive value of t, we have
`t=2pmsqrt(104) =12.2 s`
so, we get,
`D=500+20t=500+20xx12.2=744m`.