A projectle is grojected in the upward direction maging an agnle of ` 60^@` will horizontal derection with a velocity of ` 147 ms^(-1)` . The the time after which its inclination with the horizontal is ` 45^@` is.

Options

A. ` 2. 74 s`
B. ` 5 . 49 s`.
C. ` 10 . 09 s`
D. `15 s`.

Answer

B

Explanation

Let the body be projected from (O) with velocity (u) making an angle ` 60^@` with the horizontal and it reaches at (A) after tiem (t) with velocity (v) making an angle ` 45^@` with the horizontal direction. Resolving ` u` and ` v` into horizontal and verticla components we have
` u cos 60^@ = v cos 45^@ ` or ` v= u cos 60^@// cos 45^@`
` v sin 45^@ =u sin 60^@ -gt`
or ` t= (u sin 60^@-v sin 45^@)/g`
or ` t= (u sin 60^@- ( u cos 60^@ //cos 45^@ ) sin 45^@)/g`
` = u/g [ sin 60^@ - cos 60^@ ] = ( 147)/(9.8) [ (sqrt 3)/(9.8) - 1/2 ] = 5. 49 s`.