A particle is projected vertically upwards with a velocity of 196 m/sec. What will be its velocity at the end of 10 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
The question was asked in class test.
My doubt is from Calculus Application topic in chapter Application of Calculus of Mathematics – Class 12
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Correct answer is (a) 98 m/sec in the upward direction
The 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 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)*10 = 98 m/sec [as, g = 9.8m/sec^2]
Since the upward direction is taken as positive, hence after 10 seconds the velocity of the particle is 98 m/sec in the upward direction.