physics
posted by harsha ojha .
a man stand on the top of the building 490 m above the ground.he throws a stone horizontally with speed15 m/s.find the time taken by the stone to reach the ground &the speed with which it hits the ground

h=gt²/2
t=sqrt(2h/g)
v(y) = gt
v(x) =15 m/s
v= sqrt{v(x)²+v(y)²} 
A ball is thrown vertically upward with an
initial speed of 28 m/s. Then, 1.7 s later, a
stone is thrown straight up (from the same
initial height as the ball) with an initial speed
of 33.3 m/s.
How far above the release point will the ball
and stone pass each other? The acceleration
of gravity is 9.8 m/s^2
Answer in units of m
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