A gun kept on a straight horizontal road is used to hit a car travelling on the same road away from the gun at a uniform speed of 14.14 ms -1. The car is at a distance of 150 m from the gun when it is fired at an angle of 45° to the horizontal. What is the distance of the car from the gun when the shell hits it?

250 m

Explanation:

In projectile, range is given as;

R = (v²sin 2θ)/g

We are given θ = 45°

Thus range of shell is;

R = (v²sin (2 × 45))/g

R = v²/g

Now, distance for the shell to hit the car will be; 150 + 14.14t

This will be equal to the range of shell;

150 + 14.14t = v²/g

Where t is time of flight

Now, time of flight is given by;

t = (2vsin θ)/g

t = (2v sin 45)/g

In surf form, we have;

t = (2v/(g√2))

Simplifying further gives;

t = (v√2)/g

Plugging this value of t into the distance equation gives;

150 + 14.14(v√2)/g = v²/g

Assuming g is 10 m/s², we have;

150 + 14.14(v√2)/10 = v²/10

Multiply through by 10 to get;

1500 + 14.14(v√2) = v²

v² - 20v - 1500 = 0

From quadratic equation calculator, we have; v = 50 m/s

Thus, t = (50√2)/g

t = (50√2)/10

t = 5√2

Plugging this into the distance equation gives;

Distance = 150 + 14.14(5√2) = 250 m