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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 mThe question was posed to me during an interview.My question is taken from Calculus Application in chapter Application of Calculus of Mathematics – Class 12

Answer»

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.


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