mm. . for ii) 252km/h = 252000m / 3600s = 70m/s . . . there is a quicker way , its 252km/h divide by 3.6 = 70m/s . . ( dont ask me why divide by 3.6 lol . . i just know it that way lol )
for iii) since horizontal component is constant , dont worry about it. Thus theres only gravity force . F = mg . . . . a = g . . . . they want the time , and we are provided with the distance, using the relation a = dx/dt . . . dx /dt = g . . . intergrate dx = intergrade g dt . . . x = gt + C . . . when x = -270 . . t = 0 sub into the equation . C = -270 . . . x = gt - 270 . . . soo now we sub g = 10 . . . x = 10t -270 . we want to know when it hits the water . sub when x = 0 , find t . . . . t = 27. . . soo after 27 second .
iv) now we know that the thingy will travel 27s before it hits the water, and the horizontal velocity we calculated in ii) is 70m/s . . . thus this thingy will travel 27 times 70 . . . 1890m horizontally before it hits the water. and the sailor its moving 1m/s ( 3.6 divide by 3.6 ), thus in this 27 secound , the sailor moved 27m. Lets assume that it perfectly landed on the sailor . then 1890 = d + 27 . . . but they said within 30 metres , soo think of it is a circle around the sailor . . thus the shortest path will be . . . 1890 = d + (27-30) (30 is radius . . within 30m) and the other will be 1890 = d + (27+30) . . . thus d have to be ranged 1833 to 1893m . . .
btw this question dont have i) in it . . @@
Just sharing my thoughts , Peace
