A particle is projected vertically upwards with a velocity of 196 m/sec. What will be its height from the point of projection after 12 sec?

(a) 1646.2 m

(b) 1645.4 m

(c) 1644.2 m

(d) 1646.4 m

The question was posed to me during an interview.

My question is taken from Calculus Application in chapter Application of Calculus of Mathematics – Class 12

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Right answer is (a) 1646.2 m

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)

Since x = 0, when t = 0, hence from (2) we get,

0∫^x dx = 0∫^x (u – gt)dt

Or x = ut – (1/2)gt^2 ……….(3)

Let, x = h when t = 12; then from (3) we get,

h = 196*12 – (1/2)*9.8*(12*12) [as, g = 9.8m/sec^2]

= 1646.4 m.