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A particles is projected with speed u at...

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`)

A

`u//g`

B

`u//gsqrt(u^(2)-2gH)`

C

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

D

`2u//g`

Text Solution

Verified by Experts

The correct Answer is:
D
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`
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