1.

A particle is projected vertically upwards with a velocity of 196 m/sec. How much will be the greatest height?(a) 1930 m(b) 1960 m(c) 1990 m(d) 1995 mThe question was asked in a national level competition.My query is from Calculus Application in division Application of Calculus of Mathematics – Class 12

Answer»

Correct choice is (B) 1960 m

For explanation I would SAY: 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)

Again, (1) can be written as,

dv/dx*dx/dt = -g

Or v(dv/dx) = -g……….(4)

Since v = u, when x = 0, hence, from (4) we get,

u∫^vvdv = -g 0∫^xdx

Or v^2 = u^2 – 2gx……….(5)

If H m be the greatest height of the particle, then v = 0, when x = H; hence, from (5) we get,

0 = u^2 – 2gH

Or H = u^2/2g

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

= 1960 m.


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