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
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience
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.