The correct option is (a) 40 sec
For explanation: 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)
If T(≠0) be the total time of flight, then x = 0, when t = T; hence, from (3) we get,
Or 0 = uT – (1/2)gT^2
Or gT = 2u
Or T = 2u/g = 2(196)/9.8 [as, g = 9.8m/sec^2]
= 40 sec.