A particles is projected with speed `u` at an angle `theta` with horizontal particle explodes at highest point of its path into two equal fragments one of fragment moving up straighht with a speed `u`. The difference in time in which the particles fall on ground. (Assume it explodes at height `H`)

Options

A. `u//g`
B. `u//gsqrt(u^(2)-2gH)`
C. `1/(2g)sqrt(u^(2)+2gH)`
D. `2u//g`

Answer

D

Explanation

Maximum height velocity of particle `ucostheta`
`mucostheta=m/2v_(1)+0`
`v_(1)=2ucostheta`
`m/2u=m/2v_(2)`
`v_(2)=u`
Time of flight for particvle (1) is `T_(1)`
For particle (2) time of flight `T_(2)`
`uT_(1)=1/2gT_(1)^(2)=-H`
`T_(1)=(u+sqrt(u^(2)+2gH))/g`
`-v_(2)T_(2)-1/2gT_(2)^(2)=-H`
`T_(2)=(-u+sqrt(u^(2)+2gH))/g`