A small object is thrown at an angle `45^@` to the horizontal with an initial velocity `v_0` The velocity is averaged for first `sqrt2` s and the magnitude of average velocity comes out to be same as that of initial velocity, i.e. `|v_0|` . The magnitude `|v_0|` will be take, `g = 10 m//s^2` )

Options

A. `3m//s`
B. `3sqrt2`m/s
C. 4m/s
D. 5m/s

Answer

D

Explanation

Let object is at B(x, y) after t ` = sqrt(2) s `
Then ` x = u_x xx t = v_0 cos 45^@ xx sqrt2 = v_0`
and ` y = u_(y) t - 1/2 a_(y)t^2 = v_0 (sin 45^@ ) sqrt(2) - 1/2 (10) xx (sqrt2)^2 `
` = (v_0 - 10) m `
Displacement OB of particle is
`OB = sqrt(OA^2 + AB^2 ) = sqrt(v_0^2 + (v_0 -10)^2)`
so , `v_(avg) = (OB)/(t) = v_0 " ro " OB = v_0t`
`rArr sqrt(v_0^2 + (v_0 - 10)^2) = v_0 sqrt2 rArr v_0^2 + (v_0 - 10)^2 = 2v_0^2`
`rArr v_0 -10 = pm v_0 " " 2v_0 = 10 rArr v_0 = 5ms^(-1)`