A projectile is projected in the upward direction making an angle of ` 45^@` with horizontal direction with a velocity of ` 150 ms^(-1)`. Then the time after which its inclination with the horizontal is ` 30^@` is (g=10 ms^(-2)).

Options

A. ` 2.24 s`
B. ` 4.48 s`
C. ` 9. 98 s`
D. `12 . 23 s`

Answer

B

Explanation

Here, ` u = 150 ms^(-1)` , g = 10 m//s^@`
Let the body be projected from (O) with velocity (u) making an angle ` 45^2` with the horzontal and it reaches at (A) after time (t) with velocity (v) making and angle ` 30^@` with horizontal direction. Resolving (u0 and (v) into horizontal and vertical components, we have ` u cos 45^@ = v cos 30^@`
[ As in horizontal direction accleration is zero ]
` or ` v= ( u cos 45^@)/( coa 30^@)`
In vertical direction , ` sin 30^@ = u sin 45^2 - gt`
` t= 1/g [ u cos 45&@-u sin 30^@]`
`= 1/g [ u cos 45^@- - ( u cso 45^@) /( cs 30^@) xx sin 30^@]`
` = u/g [ cos 45^2 - cos 45^2 tan 30^@ ]`
` = (150)/(10) [ 1 /(sqrt 2) - 1/(sqrt 2) xx 1 /(sqrt 3)]`
` = (15)/(sqrt 2) xx (sqrt 3-1)/(sqrt 3) = (15)/(1.414) xx (( 1.632 -1))/(1 . 7 32) = 4 .48 s`.