1.

A particle is projected vertically upwards with a velocity of 196 m/sec. When will its velocity be 49m/sec in the downward direction?(a) Before 23 sec(b) After 23 sec(c) Before 25 sec(d) After 25 secI had been asked this question in homework.My query is from Calculus Application topic in section Application of Calculus of Mathematics – Class 12

Answer»

The correct choice is (d) After 25 sec

The best 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)

Let the velocity of the particle be 49m/sec in the downward direction after T seconds from the instant of its PROJECTION. Then, v = -49 m/sec, when t = T; hence, from (2) we have,

-49 = 196 – (9.8)T[as, g = 9.8 m/sec^2]

Or 9.8(T) = 245

Or T = 25

Therefore, the particle will have a velocity of 49m/sec in the downward direction after 25 seconds from the instant of its projection.


Discussion

No Comment Found

Related InterviewSolutions