An object of mass 2 m is projected with a speed of `100 ms^(-1)` at an angle,`theta = sin^(-1)(3/5)` to the horizontal. At the highest point, the object breaks into two pieces of same mass m and the first one comes to rest. The distance between the point of projection and the point of landing of the second piece (in metre) is (Given, `g = 10 ms^(-2)`).

Options

A. 3840
B. 1280
C. 1440
D. 960

Answer

C

Explanation

Horizontal range of the object is given by
`R=(U^(2) sin 2theta)/g`
At the highest point, the object breaks into 2 pieces of same and first one comes to rest Then,
`2 mu costheta = m(0) + mv`
`2 m u costheta = mv`
`v = 2u cos theta`
Total horizontal range of other part
`R/2+R=(3R)/2`
`=3/2 (u^(2)sin 2 theta)/g`
Since, `sin 2theta = 2 sintheta costheta`
`3/2 ((100)^(2) xx 2sintheta costheta)/g`
`sin theta=3/5`
` implies= cos theta= 4/5`
`=3/2xx((100)6(2)xx2xx3/5xx4/5)/10=1440`.