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While standing at the edge of the flat roof of an office building 40m high, you throw a ball upwards at 10 m/s. When does it hit the ground? what is its speed when it passes you on the way down? What is its speed when it hits the ground ?
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Added Fri, 03 Jul '15
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v = u + at
when ball goes upward,
u = 10m/s
a = -10m/s2
v = 0 at top position of ball.
0 = 10 + (-10)t
So, t=1s

Distance covered by ball in 1 sec = s = ut + 1/2 at2
    s = (10*1) + 1/2 *(-10)*1*1
    s = 5m

when ball comes downward from its top position its total distance becomes (40+5)m = 45m.
from s = ut + 1/2 at2  
    s = 45m
    u = 0
    a =10m/s2
    On calculating, t = (45*2/10)0.5 = 3 sec
    when ball hits the ground its velocity can be calculated from v2 = u2 + 2as
where, u = 0
s = 45m
a = 10m/s2
So. v = (2*45*10)0.5 = 30m/s

Now,
when the ball passes the man,
distance covered by it from top, s= 5m
       u = 0
     a = 10m/s2
from v = u + at,

v = 10m/s
Edit
Added Fri, 03 Jul '15
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