Saved Bookmarks
| 1. |
A particleprojected withvelcoityv_(0)strikesat rightalpha planethroughthepointof projection . |
|
Answer» Solution :Let `ALPHA` be the angle betweenthe velocityof projectionand the inclined plane . `v_(ax) = v_(0) cos alpha , v_(AY) = v_(0) sin alpha ` ` a_(x) = g beta , a_(y) = - g cos beta` `rArr V_(x) (t) = v_(0) cos alpha- g sin beta t` At the pointof impact ` v_(x) rArrt= (v_(0) cos alpha)/(g sin beta)""....(i)` Also y at the pointiszero . `rArr v_(0) sin alpha - (1)/(2) g cos betat^(2) = 0 rArr t = (2v_(0) sin alpha)/(g cos beta)""....(ii)` From(i) and (ii) `(v_(0) cos alpha)/(g sin beta) = (2v_(0) sin alpha)/(g cos beta)` `tan alpha =(1)/(2) cot beta` `x = v_(0) = cos (alpha + beta) t` `= v_(0)[cos alpha cos beta - sin alpha beta] (v_(0)cos alpha)/( g sin beta) = (V_(0)^(2))/(g)[cos^(2) alpha cot beta - sinalpha cos alpha]` ` = (v_(0)^(2))/(g)[((2)/(sqrt(4+cot^(2)beta)))^(2) cot beta - (cot beta)/(sqrt(4+cot^(2) beta))(2)/(sqrt(4+cot^(2)beta))]` (using`tan alpha = (1)/(2) cot beta`) = `(v_(0)^(2))/(g) (2 cotbeta)/(4+cot^(2) beta)` From FIGURE `therefore y = x tan beta = (v_(0)^(2))/(g).(2cot beta)/(4+cot^(2) beta) tan beta rArr y = (2V_(0)^(2))/(g(4+ cot^(2)beta))` 
|
|