A particle is projected vertically upwards with a velocity of 196 m/sec. What will the time of rise?

(a) 10 sec

(b) 20 sec

(c) 30 sec

(d) 40 sec

This question was posed to me by my college director while I was bunking the class.

This key question is from Calculus Application topic in section Application of Calculus of Mathematics – Class 12

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The correct answer is (b) 20 sec

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)

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)

Let, t1 be the time of rise of the particle; then v = 0, when t = t1.

Thus, from (2) we get,

0 = u – gt1

As, g = 9.8m/sec^2

Or t1 = u/g = 196/9.8 = 20 sec.