Correct choice is (b) 1960 m
For explanation I would say: Let the particle projected from A with velocity u = 196 m/sec and P be its position at time t where, AP = x m, if v m/sec be its velocity at P,then the equation of motion of the particle is,
dv/dt = -g [where, v = dx/dt] ……….(1)
Since v = u, when t = 0, hence from (1) we get,
u∫^vdv = -g 0∫^tdt
Or v – u = -gt
Or dx/dt = u – gt ……….(2)
Since x = 0, when t = 0, hence from (2) we get,
0∫^x dx = 0∫^x (u – gt)dt
Or x = ut – (1/2)gt^2 ……….(3)
Again, (1) can be written as,
dv/dx*dx/dt = -g
Or v(dv/dx) = -g ……….(4)
Since v = u, when x = 0, hence, from (4) we get,
u∫^vvdv = -g 0∫^xdx
Or v^2 = u^2 – 2gx ……….(5)
If H m be the greatest height of the particle, then v = 0, when x = H; hence, from (5) we get,
0 = u^2 – 2gH
Or H = u^2/2g
= (196*196)/(2*9.8) [as, g = 9.8 m/sec^2]
= 1960 m.