A heavy particle is usspended by a 1.5 m long string . It is given a horizontal velocityof `sqrt(57) m/s.` a. Find the angle made by the string with the upward vertical, when it becomes slack. B. Find the speed of the particle at this instant. c.Find the maximum height reached by the particle over the point of suspension. Take `g=10 m/s^2`

Answer

A, B, C

Explanation

`L=1.5m, u=sqrt57 m/sec`
a. `mgcostheta=(mv^2)/L`
`v^2=Lgcostheta`
change in K.E.= work done
` 1/2 mv^2-1/2mu^2=mgh`
`rarr v^2-57=-2xx1.5g(1+costheta)` ……….ii
`rarr v^2=57-3g(1+costheta)`

Putting the value of v from equaltion i.
`15 cos theta-57-3g(1+costheta)`
`rarr 15 costheta=57-30-30 costheta`
`rarr 45costheta=27`
`rarr costheta =3/5`
`rarr theta=cos^-13/5=53^0`
b. `v=sqrt(57-3g(1+costheta))` from equation i
`sqrt9=e/sec.
c. as the string become slack at point B,the particle will start makign projection tile motion
`H=OF+DC`
`=1.5costheta+(u^2sin^2theta)/(2g)`
`(1.5)xx3/5xx(9xx(0.8)^2)/(2xx10)`
`1.2m`