Physics
posted by Anonymous .
Standing on the top ledge of a 55 m high building you throw a ball straight up with an initial speed of 45 m/s. How long to the nearest second, does it take to hit the ground?
I got 10 seconds for the first question.
How do I go about answering this one?
If the building in the previous problem is 48 meters high and you throw the ball up at 30 m/s, how high to the nearest meter does it go?

well, once it stops at the top, it is a simple fall from there.
How high above the building does it go and how long to the top?
(45 m/s is quite a throw straight up)
m g h = (1/2) m v^2
h = (1/2) (45)^2/9.81 = 103 m
average speed up = 45/2 = 22.5 m/s
so time in air up =103/22.5 = 4.58 s upward
now it falls from 55+103 = 158 m
d = (1/2) g t^2
158 = 4.9 t^2
t = 5.68 s
total time in air = 4.58 + 5.68 = 10.3 s
the way I did it the second question is really easy 
The correct answer was 94 meters for the second part of the question. That's what I'm unsure as to how to get it...
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