Question

plz solve the problems

Answer 1

ANSWER :-


$solution$


As at the time of firing of the shell, the particle was at C and the shell collides with it at B, therefore the number of the revolution completed by the particle is odd multiple of half .,(2n - 1)/2, where n is an integer.


Let T be the time period of the particle, then

T = 2πr/v = 2 * 3.14 * 20/31.4

= 4 second

if t be the time of the flight of the shell , then

t = time of [ (2n - 1)] second

= (2n - 1 ) /2 * 4

= 2(2n - 1) second

for a projectile, the time of flight is given by

t = 2u sin¢/g

Hence,


2u sin¢/g = 2(2n - 1) .........(1)



the range of the projectile is given by

R = u²sin2¢/g

Hence,

u² sin2¢/g = 20√3 ........…(2)


From equation (1) and (2)

tan¢ = (2n - 1)²/√3

=> ¢= 30°

For ¢ to be smallest , n = 1

so,

tan¢ = (2n - 1)²/√3

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