A projectile is projected in the upward direction making an angle of ` 60^@` will horizontal direction with a velocity of ` 147 ms^(-1)` . The the time after which its inclination with the horizontal is ` 45^@` is.

Options

A. 15 s
B. 10.98 s
C. 5.49 s
D. 2.745 s

Answer

C

Explanation

At any points of the trajectory during projection, the horizontal component of the velocity is the same-
`u cos 60^(@) = v cos 45^(@)`
`rArr v_(x) = 147 xx 1/2 xx sqrt(2) = 147/sqrt(2) m//s`
`theta = 60^(@), u_(y) = u sin 60^(@) = (147sqrt(3))/2 m//s`
`theta = 45^(@), v_(y) = v sin 45^(@) = 147/sqrt(2) xx 1/sqrt(2) = 147/2 m//s`
Now, `v_(y) = u_(y) + a_(y)t rArr _(y) = u_(y) - "gt"`
`rArr t = (u_(y) - v_(y))/g`
`= ((147sqrt(3))/2 - 147/2)/9.8 = 147/(2 xx 9.8) (sqrt(3)-1)`
`rArr t = 5.475 sec = 5.49 sec`