A particle is projected vertically upwards with a velocity of 196 m/sec. How many times will it attain a height of 1254.4 m after projection?
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This question is from Calculus Application topic in portion Application of Calculus of Mathematics – Class 12
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Correct option is (c) 2
To explain: 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)
Suppose the particle attains the height of 1254.4 m after t2 seconds from the instant of projection.
Then, x = 1254.4, where t = t2; hence, from (3) we get,
1254.4 = 196t2 – (1/2)(9.8)t2^2 [as, g = 9.8 m/sec^2]
t2^2 – 40t2 + 256 = 0
Or (t2 – 8)(t2 – 32) = 0
Or t2 = 8, 32
Clearly, the particle attains the height of 1254.4m twice; once after 8 seconds from the instant of its projection during its upward motion and next, after 32 seconds from the same instant during its motion in the downward direction.