A bomb is released from a plane which is diving at an angle of 30 degrees from the vertical . The plane is at a height of 1000m when the bomb is released . The bomb hits the ground 10s later . What is the speed of the plane when the bomb is released .

My attempt :

$\displaystyle s=ut+\frac{1}{2}at^2$

$\displaystyle 1000=usin30(10)+frac{1}{2}(9.81)(10)^2$

$\displaystyle

u= 101.9 mms^{-1}

$

But it is wrong .. Wonder where my mistake is .. THanks for helping me out .