A projectile is thrown in the upward direction making an angle of `60^@` with the horizontal with a velocity of `140 ms^-1`. Then the time after which its velocity makes an angle `45^@` with the horizontal is (Acceleration due to gravity `g=10 ms^-2`)

Options

A. 0.5124 s
B. 51.24 s
C. 5.124 s
D. 512.4 s

Explanation

Given angle of projection, `theta=60^@` and velocity , `mu=140 ms^-1`
This velocity have divided into two components, horizontal component, `mu_x=mu cos60^@=140 cos 60^@` and vertical component `mu_y= mu sin60^@=140 sin 60^@` Let after time t, the inclination of particle with horizontal be `45^@` and at time t velocity along `x=v_x ` and along `y=v_y`.
Now, `tan45^@=v_y/v_x`
`=v_x=v_y`
Since the horizontal component of velocity remains constant, i.e.,
`v_x=mu_x=140 cos60^@`
and `v_y=mu_y- gt`
`140 cos60^@=140 sin60^@ -10t[becausev_y=v_x]`
`140 times1/2=140times sqrt3/2-10t`
`70=70sqrt3-10t`
`10t=70sqrt3-70`
`t=(70(sqrt3-1))/10`
`=7times(0.7320)=5.124s`