A particle is projected vertically upwards with a velocity of 196 m/sec. What will the time of rise?

A particle is projected vertically upwards with a velocity of 196 m/sec. What will the time of rise? Correct Answer 20 sec

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 ……….(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)gt2 ……….(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 v2 = u2 – 2gx ……….(5) Let, t1 be the time of rise of the particle; then v = 0, when t = t1. Thus, from (2) we get, 0 = u – gt1 As, g = 9.8m/sec2 Or t1 = u/g = 196/9.8 = 20 sec.