A particle projected withvelocity v_(0), strikes at right angles alpha plane through the point of projection and of inclination beta with the horizontal. Find the height of the point struck, from horizontal plane through the point of projection.

A particle projected withvelocity v_(0), strikes at right angles alpha

1.	A particle projected withvelocity v_(0), strikes at right angles alpha plane through the point of projection and of inclination beta with the horizontal. Find the height of the point struck, from horizontal plane through the point of projection.
Answer» Solution :Let `alpha` be the ANGLE between the velocity of projection and the inclined plane. `v_(ox.)=v_(0)cos alpha, v_(oy.)=v_(0)sin alpha` `a_(x.)=-g beta""a_(y.)=-g cos beta` `implies v_(x.)(t)=v_(0)cos alpha-g sin beta t` At the point of impact `v_(x.)=0impliest=(v_(0)cos alpha)/(g sin beta)`........i At y. at the point is zero `impliesv_(0)sin alpha t -1/2 g cos beta t^(2)=0impliest=(2v_(0)sin alpha)/(g cos beta)`......ii From IA nd ii `(v_(0)cos alpha)/(g sin beta)=(2v_(0)sin alpha)/(g cos beta)` `tan alpha=1/2cot beta` .................iii `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-sin alpha cos alpha]` `=(v_(0)^(2))/g[(2/(sqrt(4+cot^(2)beta)))^(2)cotbeta-(cot beta)/(sqrt(4+cot^(2)beta))2/(sqrt(4+cot^(2)beta))]` (using `tan alpha=1/2cot beta)=(v_(0)^(2))/g .(2cot beta)/(4+cot^(2)beta)` From figure `:.y=x tan beta=(V_(0)^(2))/g . (2 cot beta)/(4+cot^(2)beta) tan beta impliesy=(2V_(0)^(2))/(g (4+cot^(2)beta))`