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A particle is projected with speed u in ...

A particle is projected with speed u in air at an angle `theta` with the horizontal. The particle explodes at the highest point of its path into two equal fragment, one of the ftagments movingg up straight with a speed u. The difference in times in which the two particles fall on the ground is (Assume it is at a height H at the time of explosion):

A

`(2u)/(g)`

B

`(u)/(g) sqrt(u^(2) - 2 gH)`

C

`(1)/(2g) sqrt(u^(2) + 2gH)`

D

`(2)/(g) sqrt(u^(2) + 2gH)`

Text Solution

Verified by Experts

The correct Answer is:
A

`vec(P)_(i) = vec(P)_(f)`
`m u cos theta hat(i) = - (m)/(2) u hat(j) + (m)/(2) vec(V)`
`vec(V) = 2 u cos theta hat(i) + u hat(j)`

`T_(1) =` Time to go from A to B + Time to go from B to C.
`= 2 (u)/(g) +` Time to go from B to C.

`T_(2) =` time to go from B to C
`Delta T = 2 (u)/(g)`
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