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a bomber flying horizontally at a speed of 300km/hr at an altitude of 150m releases a bomb targeting a ship moving in the same direction as the plane at a constant speed of 10m/s. how far from the ship should it release the bomb to hit it? also determine the angle of sight.
sourabh
on 2012-10-28 09:30:00Key concepts: Relative velocity; vertical and horizontal components of velocity. Draw a schematic diagram first, it always helps in understanding. velocity of plane = 83.33m/s velocity of ship = 10m/s Now, relative speed of plane wrt ship = 83.33-10= 73.33 m/s. Now, the time required for bomb to travel vertical height of 150m and the required distance btwn ship and plane would be the same. vertical component of velocity of bomb,uy = 0 vertical distance, sy = 150m acceleration, g = 9.81m/s*2 so, sy = uy*t + 0.5*g*t^2 hence, t = 5.53 s again, if reqd distance be sx, then sx = vx*t or, sx = (73.33*5.53) m or, sx = 405.51 m now, angle of sight be p so,tan p = (vertical distance/horizontal dist) or, p= atan(150/405.51) or, p= 20.3 degrees approx. CHECK ANSWER IF AVAILABLE AND INDICATE IF SOLUTION IS CORRECT OR WRONG