Question

A particle is projected vertically upwards with a velocity of 196 m/sec. What will be its velocity at the end of 30 seconds?

(a) 98 m/sec in the upward direction

(b) 98 m/sec in the downward direction

(c) 99 m/sec in the upward direction

(d) 99 m/sec in the downward direction

I have been asked this question in examination.

The origin of the question is Calculus Application topic in chapter Application of Calculus of Mathematics – Class 12

Answer 1

Correct answer is (b) 98 m/sec in the downward direction

The best I can explain: 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 v1 be the velocities of the particle after 10 seconds from the instant of projection.

Then v = v1 when t = 10; hence from (2) we get,

v1 = 196 – (9.8)*30 = -98 m/sec  [as, g = 9.8m/sec^2]

Since the upward direction is taken as positive, hence after 30 seconds the velocity of the particle is -98 m/sec in the upward direction. So, the velocity will be 98 m/sec in the downward direction.