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?

Category: QuestionsA 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?
Editor">Editor Staff asked 11 months ago

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
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

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.