एक कण को क्षैतिज से `30^@` के कोण पर 20 m/s के वेग से प्रक्षेपित किया जाता है : `(g = 10 m//s^2)`
पथ के उच्चतम बिन्दु पर वक्रता त्रिज्या ज्ञात कीजिए ।

Options

A. 30 m
B. 20 m
C. 40 M
D. 50 m

Answer

A

Explanation

(i) `x = u cos theta t`
`= 20 xxsqrt3/2xxt=10 sqrt3m`
`y = u sintheta t -1/2xx10 xxt^2`
`= 20 xx1/2 xx(A) - 5 (A)^2 = 5m`
स्थिति सदिशा `= vecr = 10 sqrt3hati+5hatj, |vecr|=sqrt((10sqrt3)^(2)+5^2)`
(ii) `v_x= 10 sqrt3hati`
`v_y = u_y+a_yt = 10 - "g t "=0`
`:.vecv=10sqrt3hati, " "|vecv|=10sqrt3`
`vecv.vecr=(10sqrt3hati)(10sqrt3hati+5hatj)=300`
`vecv.vecr=|vecv||vecr|costheta`
`impliescos theta= (vecv.vecr)/(|vecv||vecr|)=300/(10sqrt3sqrt325)implies theta= cos^(-1)(2sqrt3/13)`
(ii) `v_x=10sqrt3hati`
`v_y=u_(y) +a_(y)t = 10 - gt = 0`
`:. vecv=10 sqrt3hati`
`:. vecv = 10sqrt3hati`
`vevv.vecr=(10sqrt3.hati)(10sqrt3 hati+5hatj)=300`
`vecv.vecr=|vecv||vecr|costheta`
`implies cos theta=(vecv.vecr)/(|vecv||vecr|)=300/(10sqrt3sqrt325)impliestheta=cos^(-1)(2sqrt((3)/(13)))`