Question

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 sec

I had been asked this question in homework.

My query is from Calculus Application topic in section Application of Calculus of Mathematics – Class 12

Answer 1

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.