A ball is dropped onto a step at a point P and rebounds with velocity `u_@` at an angle of `45^@ `with the vertical. The value of `u_0` knowing that just before the ball hits the point Q its velocity forms an angle `30^@` with the vertical is: `(g=10m//s^2)`

Options

A. (a)2 m/s
B. (b)1 m/s
C. (c)`sqrt(2)` m/s
D. (d)`sqrt(3)` m/s

Answer

C

Explanation

At point Q `V_("horizontal")=u_(0) sin 45^(@)`
`V_("vertical")=sqrt((u_(0) cos 45^(@))^(2)+2gh) ("Down") tan30^(@)=V_("horizontal")/(V_("vertical"))=(u_(0) sin 45^@)/(sqrt(u_(0) cos 45^(@))^(2)+2gh)`
`1/sqrt3=(u_(0)//sqrt2)/(sqrt((u_(0)^(2)/(2)+2xx 10 xx 0.1)) rArr 1/3= (u_(0)^2//2)/(u_(0)^(2)/2+2) u_(0)=sqrt2`