physics
posted by rista on .
a ball is projected horizontally from the edge of a table that is 0.433m high, and it strikes the floor at a point 1.84m from the base of the table. the acceleration of gravity is 9.8m/s^2.what is the initial speed of the ball? how high is the ball above the floor when its velocity vector makes a 18.9486 angle with the horizontal?
i got initial speed that is 6.13 m/s. but i don't know to find another question.. please help...

The time it takes to hit the ground is
t = sqrt(2H/g) = 0.2973 s
The speed when it leaves is the constant horizontal component
Vx = 1.84/0.306 = 6.19 m/s
When the velocity vector angle is 18.9486 degrees below horizontal, the ratio Vy/Vx is the tangent of 18.9486 degrees, which is 0.3433
Therefore Vy = 2.13 m/s at that time.
The time after leaving is
t' = 2.13/g = 0.217 s
Use that time to calculate the distance it has fallen, and from that, the height above the floor. 
hey i tried bt i didn't got the answer.. can u help more to get answer.. please