Two bodies were thrown simultaneously from the same point, one, straight up, and the other, at an angle of `theta=60^@` to the horizontal. The initial velocity of each body is equal to `v_0=25m//s`. Neglecting the air drag, find the distance between the bodies `t=1.70s` later.

Options

A. `20m`
B. `18m`
C. `22m`
D. `24m`

Answer

C

Explanation

At `t=0,vecr_(rel)=0`
`vecu_(1)-vecu_(2)=(v_(0)-v_(0)sin theta)hatj-v_(0)cos thetahati , veca_(rel)=vec0`
After time `t=1.7` second , Seperation is `vecs_(rel=vecr_(1)-vecr_(2)`
`vecs_(rel)=vecu_(rel) "t"+1/2a_(rel)t^(2),vecs_(rel)=(vecu_(1)-vecu_(2))t`
`vecs_(rel)={(v_(0)-v_(0)sin theta)hatj-v_(0)cos theta hati}t`
`vecs_(rel)=sqrt{{(v_(0)-v_(0)sin theta) "t"}+(-v_(0)t cos theta)^(2)`
`=v_(0)tsqrt(2(1-sin theta))=22m`