Answer to Question #94683 in Mechanics | Relativity for sameer shahzad

Question

A gun on shore fires a shot at a ship which is heading directly towards the gun at a speed of 40 km/h. At instant of firing,the distance to the ship is 15000 m . The muzzle speed of the shot is 700 m/s.pretend that there is no air resistance.(A)What is the required elevation angle for the gun ?Assume g=9.8 m/s^2 (B)what is the time interval between firing and impact?

Answer 1

a) For the projectile motion, range of the fire is determined by:

$$L=\frac{v^2\sin2\phi}{g}$$

So,

$$\phi=\frac{\sin^{-1}(\frac{Lg}{v^2})}{2}=\frac{\sin^{-1}(\frac{15000*9.8}{700^2})}{2}=8.7^{\circ}$$

b)

The total flight time is:

$$t_{tot}=2t_{up}=2\frac{v_{vert}}{g}=2\frac{v*\sin\phi}{g}=2\frac{700*\sin8.7^{\circ}}{9.8}=21.6s$$

Answer: a) 8.7 degrees, b) 21.6 seconds